Linear and Quadratic Equations in One Variable

Mathematics 17

Institute of Mathematics, University of the Philippines-Diliman

Lecture 7

Math 17 (UP-IMath) Equations Lec 7 1 / 28

Outline

Equations in One Variable

1 Introduction: Equations as Statements

2 Solving Linear and Quadratic Equations in One VariableLinear EquationsQuadratic EquationsVerbal Problems

Math 17 (UP-IMath) Equations Lec 7 2 / 28

Equations as Statements

Definition

An equation is a statement saying that two expressions are equal.

Examples:

1. 2x+ 1 = x 72.

1

2 z =2

4 2z3.

2

y + 1=

1

y

Math 17 (UP-IMath) Equations Lec 7 3 / 28

Equations as Statements

Definition

An equation is a statement saying that two expressions are equal.

Examples:

1. 2x+ 1 = x 72.

1

2 z =2

4 2z3.

2

y + 1=

1

y

Math 17 (UP-IMath) Equations Lec 7 3 / 28

Equations as Statements

Definition

If an equation holds true for every permissible value (in R), then theequation is called an identity.

Example:1

2 z =2

4 2z is true for any z 6= 2.

Definition

If an equation is never true for any permissible value (in R), then theequation is called a contradiction.

Example: 1 2x = 3 2x is never true for any value of x R.

Math 17 (UP-IMath) Equations Lec 7 4 / 28

Equations as Statements

Definition

If an equation holds true for every permissible value (in R), then theequation is called an identity.

Example:1

2 z =2

4 2z is true for any z 6= 2.

Definition

If an equation is never true for any permissible value (in R), then theequation is called a contradiction.

Example: 1 2x = 3 2x is never true for any value of x R.

Math 17 (UP-IMath) Equations Lec 7 4 / 28

Equations as Statements

Definition

If an equation holds true for every permissible value (in R), then theequation is called an identity.

Example:1

2 z =2

4 2z is true for any z 6= 2.

Definition

If an equation is never true for any permissible value (in R), then theequation is called a contradiction.

Example: 1 2x = 3 2x is never true for any value of x R.

Math 17 (UP-IMath) Equations Lec 7 4 / 28

Equations as Statements

Definition

If an equation holds true for every permissible value (in R), then theequation is called an identity.

Example:1

2 z =2

4 2z is true for any z 6= 2.

Definition

If an equation is never true for any permissible value (in R), then theequation is called a contradiction.

Example: 1 2x = 3 2x is never true for any value of x R.

Math 17 (UP-IMath) Equations Lec 7 4 / 28

Equations as Statements

Definition

If an equation holds true for some value(s), then the equation is aconditional equation.

Examples:

1 2x+ 1 = x 7 holds if x = 82 x2 3x 4 = 0 is true if x = 1 or x = 4

Math 17 (UP-IMath) Equations Lec 7 5 / 28

Equations as Statements

Definition

If an equation holds true for some value(s), then the equation is aconditional equation.

Examples:

1 2x+ 1 = x 7 holds if x = 8

2 x2 3x 4 = 0 is true if x = 1 or x = 4

Math 17 (UP-IMath) Equations Lec 7 5 / 28

Equations as Statements

Definition

If an equation holds true for some value(s), then the equation is aconditional equation.

Examples:

1 2x+ 1 = x 7 holds if x = 82 x2 3x 4 = 0 is true if x = 1 or x = 4

Math 17 (UP-IMath) Equations Lec 7 5 / 28

Solution Set of an Equation

Definition

1 A solution (or root) of an equation is a value of the variable thatmakes the equation true.

2 The solution set of an equation is the set of all solutions of theequation.

Unless specified, the solution sets that we consider are subsets of R.

GOAL:

To find the solution set of a given equation

Math 17 (UP-IMath) Equations Lec 7 6 / 28

Solution Set of an Equation

Definition

1 A solution (or root) of an equation is a value of the variable thatmakes the equation true.

2 The solution set of an equation is the set of all solutions of theequation.

Unless specified, the solution sets that we consider are subsets of R.

GOAL:

To find the solution set of a given equation

Math 17 (UP-IMath) Equations Lec 7 6 / 28

Linear Equations in x

Definition

Linear equations are equations which upon algebraic manipulation involvesonly polynomials with degree 1.

Example: x 5 = 2x+ 7

Math 17 (UP-IMath) Equations Lec 7 7 / 28

Solving Linear Equations in x

Processes in solving equations:

1 Transposing terms (Additive Property of Equality)

2 Simplifying terms

3 Multiplying/dividing both sides of the equation by a nonzero realnumber (Multiplicative Property of Equality)

Any linear equation in x can be written, using the processes above, in theform

ax+ b = 0,

where a, b R and a 6= 0.

Its solution set is

{ ba

}.

Math 17 (UP-IMath) Equations Lec 7 8 / 28

Solving Linear Equations in x

Processes in solving equations:

1 Transposing terms (Additive Property of Equality)

2 Simplifying terms

3 Multiplying/dividing both sides of the equation by a nonzero realnumber (Multiplicative Property of Equality)

Any linear equation in x can be written, using the processes above, in theform

ax+ b = 0,

where a, b R and a 6= 0.

Its solution set is

{ ba

}.

Math 17 (UP-IMath) Equations Lec 7 8 / 28

Solving Linear Equations in x

Processes in solving equations:

1 Transposing terms (Additive Property of Equality)

2 Simplifying terms

3 Multiplying/dividing both sides of the equation by a nonzero realnumber (Multiplicative Property of Equality)

Any linear equation in x can be written, using the processes above, in theform

ax+ b = 0,

where a, b R and a 6= 0.

Its solution set is

{ ba

}.

Math 17 (UP-IMath) Equations Lec 7 8 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.

Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)

5x 5 2x 7 = 0 (Addition Property of Equality)3x 12 = 0 (Simplification)

x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)

x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Solving Linear Equations in x

Example: Solve for the value of x in 5x 5 = 2x+ 7.Solution:

5x 5 = 2x+ 7 (Given)5x 5 2x 7 = 0 (Addition Property of Equality)

3x 12 = 0 (Simplification)x = 123 (solution of ax+ b = 0)

Checking: if x = 4, 5(4) 5 = 15 and 2(4) + 7 = 15.

Solution set : {4}

Math 17 (UP-IMath) Equations Lec 7 9 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.

3x(2 x) + 7x+ 3 = 5x 3x2 + 16x 3x2 + 7x+ 3 5x+ 3x2 1 = 0

8x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x

3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2

+ 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3

5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x

+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2

1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1

= 0

8x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 0

8x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x

+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2

= 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x

= 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2

x = 28= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x =

28= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8

= 14

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Example: Solve for x in 3x(2 x) + 7x+ 3 = 5x 3x2 + 1.3x(2 x) + 7x+ 3 = 5x 3x2 + 1

6x 3x2 + 7x+ 3 5x+ 3x2 1 = 08x+ 2 = 0

8x = 2x = 2

8= 1

4

Solution set:

{14

}

Math 17 (UP-IMath) Equations Lec 7 10 / 28

Quadratic Equations in x

Definition

A quadratic equation is an equation which upon algebraic manipulationinvolves only polynomials of degree 2.

Example: x2 + 3x = 10

x2 + 3x 10 = 0

Any quadratic equation in x can be written in the form

ax2 + bx+ c = 0,

where a, b, c R and a 6= 0.

Math 17 (UP-IMath) Equations Lec 7 11 / 28

Quadratic Equations in x

Definition

A quadratic equation is an equation which upon algebraic manipulationinvolves only polynomials of degree 2.

Example: x2 + 3x = 10

x2 + 3x 10 = 0

Any quadratic equation in x can be written in the form

ax2 + bx+ c = 0,

where a, b, c R and a 6= 0.

Math 17 (UP-IMath) Equations Lec 7 11 / 28

Quadratic Equations in x

Definition

A quadratic equation is an equation which upon algebraic manipulationinvolves only polynomials of degree 2.

Example: x2 + 3x = 10 x2 + 3x 10 = 0

Any quadratic equation in x can be written in the form

ax2 + bx+ c = 0,

where a, b, c R and a 6= 0.

Math 17 (UP-IMath) Equations Lec 7 11 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0

(x+ 5)(x 2) = 0x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0

or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0

x = 5 or x = 2Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5

or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.

Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Recall: If ab = 0, then a = 0 or b = 0.

Example: Solve for x in x2 + 3x = 10.

x2 + 3x 10 = 0(x+ 5)(x 2) = 0

x+ 5 = 0 or x 2 = 0x = 5 or x = 2

Checking:If x = 5, (5)2 + 3(5) = 10.If x = 2, (2)2 + 3(2) = 10.Solution set: {5, 2}

Math 17 (UP-IMath) Equations Lec 7 12 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.

6x2 11x 10 = 0(3x+ 2)(2x 5) = 0

3x+ 2 = 0 or 2x 5 = 0x = 2

3or x =

5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 0

3x+ 2 = 0 or 2x 5 = 0x = 2

3or x =

5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0

or 2x 5 = 0x = 2

3or x =

5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2

Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

Solving Quadratic Equations in x via Factoring

Example: Solve for x in 6x2 11x 10 = 0.6x2 11x 10 = 0

(3x+ 2)(2x 5) = 03x+ 2 = 0 or 2x 5 = 0

x = 23

or x =5

2Checking:

If x = 23 , 6(23)2 + 11(23) 10 = 0.If x = 52 , 6(

52)

2 + 11(52) 10 = 0.

Solution set:{23 , 52}

Math 17 (UP-IMath) Equations Lec 7 13 / 28

What if ax2 + bx+ c is not factorable?

Note: If u2 = d, then u = d.

For example, if u2 = 9, then u = 3.

Recall: Perfect square trinomial a2 2ab+ b2

Math 17 (UP-IMath) Equations Lec 7 14 / 28

What if ax2 + bx+ c is not factorable?

Note: If u2 = d, then u =

d.

For example, if u2 = 9, then u = 3.

Recall: Perfect square trinomial a2 2ab+ b2

Math 17 (UP-IMath) Equations Lec 7 14 / 28

What if ax2 + bx+ c is not factorable?

Note: If u2 = d, then u = d.

For example, if u2 = 9, then u = 3.

Recall: Perfect square trinomial a2 2ab+ b2

Math 17 (UP-IMath) Equations Lec 7 14 / 28

What if ax2 + bx+ c is not factorable?

Note: If u2 = d, then u = d.

For example, if u2 = 9, then u = 3.

Recall: Perfect square trinomial a2 2ab+ b2

Math 17 (UP-IMath) Equations Lec 7 14 / 28

What if ax2 + bx+ c is not factorable?

Note: If u2 = d, then u = d.

For example, if u2 = 9, then u = 3.

Recall: Perfect square trinomial a2 2ab+ b2

Math 17 (UP-IMath) Equations Lec 7 14 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = c

x2 +b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2

+b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax

= ca

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2

+ 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax

= ca

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax

+b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2

= ca+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2

(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2

=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=

4ac+ b24a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x =

b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2a

b2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2a

b2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

ax2 + bx+ c = 0

ax2 + bx = cx2 +

b

ax = c

a

x2 + 2 b2ax = c

a

x2 + 2 b2ax+

b2

4a2= c

a+

b2

4a2(x+

b

2a

)2=4ac+ b2

4a2

x+b

2a=

b2 4ac4a2

x = b2ab2 4ac2a

x =bb2 4ac

2a

Math 17 (UP-IMath) Equations Lec 7 15 / 28

The Quadratic Formula

Let a, b, c R with a 6= 0. If ax2 + bx+ c = 0, then

x =bb2 4ac

2a.

Math 17 (UP-IMath) Equations Lec 7 16 / 28

Solve for x in 6x2 11x 10 = 0.

We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =

11121 4(6)(10)2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11

121 4(6)(10)2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11

121 4(6)(10)2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11

121 4(6)(10)2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121

4(6)(10)2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11

121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12

or x =11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12

or x =812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2

or x = 23

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solve for x in 6x2 11x 10 = 0.We expect x = 23 and x = 53 .

By the quadratic formula,

x =11121 4(6)(10)

2(6)

=11121 + 240

12

=11 1912

x =11 + 19

12or x =

11 1912

x =30

12or x =

812

x =5

2or x = 2

3

Math 17 (UP-IMath) Equations Lec 7 17 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636 4(1)(1)

2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =

636 4(1)(1)2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =6

36 4(1)(1)2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636

4(1)(1)2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636 4(1)(1)

2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636 4(1)(1)

2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636 4(1)(1)

2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Solving Quadratic Equations in x via Quadratic Formula

Example: Solve for x in x2 6x 1 = 0.

x =636 4(1)(1)

2(1)

=640

2

=6 210

2

= 310

Checking:(310)2 6(310) 1 = (9 610 + 10) (18 610) 1) = 0.

Solution set: {3 +10, 310}.

Math 17 (UP-IMath) Equations Lec 7 18 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a , the discriminant gives the

nature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions...zero, then the quadratic equation has one real solution...negative, then the quadratic equation has two imaginary solutions

that are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a ,

the discriminant gives thenature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions...zero, then the quadratic equation has one real solution...negative, then the quadratic equation has two imaginary solutions

that are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a , the discriminant gives the

nature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions...zero, then the quadratic equation has one real solution...negative, then the quadratic equation has two imaginary solutions

that are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a , the discriminant gives the

nature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions.

..zero, then the quadratic equation has one real solution.

..negative, then the quadratic equation has two imaginary solutionsthat are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a , the discriminant gives the

nature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions...zero, then the quadratic equation has one real solution.

..negative, then the quadratic equation has two imaginary solutionsthat are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

Definition

The discriminant of the quadratic polynomial ax2 + bx+ c is b2 4ac.

From the quadratic formula x = bb24ac2a , the discriminant gives the

nature of the solutions of a quadratic equation:

If b2 4ac is....positive, then the quadratic equation has two distinct real solutions...zero, then the quadratic equation has one real solution...negative, then the quadratic equation has two imaginary solutions

that are conjugates of each other.

Math 17 (UP-IMath) Equations Lec 7 19 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1)

= 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0,

so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution.

Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: 4x2 + 4x+ 1 = 0

Note: 16 4(4)(1) = 16 16 = 0, so we expect the equation to haveexactly one real solution. Indeed,

4x2 + 4x+ 1 = 0

(2x+ 1)2 = 0

2x+ 1 = 0

x = 12

Math 17 (UP-IMath) Equations Lec 7 20 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0

Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2)

= 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0,

so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions.

Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =

24 82

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =2

4 82

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24

82

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Nature of Solutions of a Quadratic Equation

A quadratic equation may have 2 distinct real solutions, 1 unique realsolution, or 2 imaginary solutions that are conjugates of each other.

Example: x2 2x+ 2 = 0Note: 4 4(1)(2) = 4 8 = 4 < 0, so we expect the equation to havetwo imaginary solutions. Indeed,

x =24 8

2

=24

2

=2 2i2

= 1 i

Math 17 (UP-IMath) Equations Lec 7 21 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Suggestions in Solving Verbal Problems

1 Read the problem carefully. Determine the quantity/ies that areknown and those that are unknown.

2 Represent an unknown quantity by a variable. If possible, express theother unknown quantities in terms of this variable.

3 If possible, draw a diagram or make a table depicting the relationshipsbehind the problem.

4 Set up an equation or equations that would relate the variables, orthose that can be deduced from the diagram or table.

5 Solve the equation, inequality or system of equations. Identifycorrectly what is being asked. Do not forget to indicate the unit ofmeasurement.

6 Verify that your answer/s indeed satisfy/ies the conditions of theproblem.

Math 17 (UP-IMath) Equations Lec 7 22 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).

Since distance=speedtime,After t hours: Plane 1 has travelled 480t mi

Plane 2 has travelled 520t miThe planes are 2000 mi apart

Hence 480t+ 520t = 20001000t = 2000

t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled

480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t mi

Plane 2 has travelled 520t miThe planes are 2000 mi apart

Hence 480t+ 520t = 20001000t = 2000

t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apart

Hence 480t+ 520t = 20001000t = 2000

t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000

t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Uniform Motion

Example: Two airplanes traveling in opposite directions left an airport atthe same time. If one plane is traveling at a constant speed of 480 mi/hrand the other is traveling at a constant speed of 520 mi/hr, how long willit take until they are 2000 mi apart?

Solution: Let t be this unknown time (in hours).Since distance=speedtime,

After t hours: Plane 1 has travelled 480t miPlane 2 has travelled 520t mi

The planes are 2000 mi apartHence 480t+ 520t = 2000

1000t = 2000t = 2

Checking: if t = 2, 480t+ 520t = 2000.

Thus, the planes are 2000 mi apart after 2 hours.

Math 17 (UP-IMath) Equations Lec 7 23 / 28

Mixture Problem

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Note:

Volume of solution 1 + Volume of solution 2= Volume of combined solution

Amount of solution Concentration (in %) of thesolute= Amount of solute in the solution

Amount of solute in solution 1 + Amount of solute insolution 2= Amount of solute in the combined solution

Math 17 (UP-IMath) Equations Lec 7 24 / 28

Mixture Problem

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Note:

Volume of solution 1 + Volume of solution 2= Volume of combined solution

Amount of solution Concentration (in %) of thesolute= Amount of solute in the solution

Amount of solute in solution 1 + Amount of solute insolution 2= Amount of solute in the combined solution

Math 17 (UP-IMath) Equations Lec 7 24 / 28

Mixture Problem

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Note:

Volume of solution 1 + Volume of solution 2= Volume of combined solution

Amount of solution Concentration (in %) of thesolute= Amount of solute in the solution

Amount of solute in solution 1 + Amount of solute insolution 2= Amount of solute in the combined solution

Math 17 (UP-IMath) Equations Lec 7 24 / 28

Mixture Problem

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Note:

Volume of solution 1 + Volume of solution 2= Volume of combined solution

Amount of solution Concentration (in %) of thesolute= Amount of solute in the solution

Amount of solute in solution 1 + Amount of solute insolution 2= Amount of solute in the combined solution

Math 17 (UP-IMath) Equations Lec 7 24 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X

x 20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:

Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X

x 20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).

The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X

x 20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X

x 20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x

20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20%

0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y

100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100

50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50%

50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100

30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100 30%

0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution:Let x be the amount of Solution X needed (in liters).The mixture problem could be presented as follows:

Amt of Sol(liters)

Acid ConcentrationAmt of Acid

(liters)

Sol X x 20% 0.20x

Sol Y 100 50% 50

Result.Sol

x+ 100 30% 0.30(x+ 100)

Thus,

0.20x+ 50 = 0.30(x+ 100)

Math 17 (UP-IMath) Equations Lec 7 25 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x

+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500

= 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 =

3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x

+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500

300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300

= 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 =

3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x

2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x

200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Solution X is 20% acid while Solution Y is 50% of the sameacid. How much of Solution X must be added to a 100 liters of SolutionY to get a solution that is 30% acid?

Solution (cont):0.20x+ 50 = 0.30(x+ 100)

10[0.20x+ 50] = 10[0.30(x+ 100)]

2x+ 500 = 3x+ 300

500 300 = 3x 2x200 = x

Checking: If x = 200, 0.20x+ 50 = 90 and 0.30x+ 30 = 90.

Hence, 200 liters of Solution X is needed.

Math 17 (UP-IMath) Equations Lec 7 26 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers.

Then,

x+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x

+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2)

+ (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2) + (x+ 4)

= (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6 = x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check: Indeed, 1 + 3 + 5 = 32.

Hence, the three consecutive odd integers are 1, 3 and 5.

Math 17 (UP-IMath) Equations Lec 7 27 / 28

Example. Find three consecutive odd numbers whose sum is the square ofthe second odd number.

Solution: Let x be the smallest of the odd numbers. Then,

x+ (x+ 2) + (x+ 4) = (x+ 2)2

3x+ 6

= x2 + 4x+ 4

0 = x2 + x 20 = (x+ 2)(x 1)

x = 2 or x = 1Since x is odd, x must be 1, and the other integers must be 3 and 5.

Check