Thinking Mathematically

William E. Blommer

James M. Lueken

Equivalence Properties of Equality

-All are quite easy to understand, but are very similar. Make sure toknow the differences of these properties.

A. Reflexive Property (of Equality)Example: The property that a = a.Hint: It’s your reflex to think a=a

B. Symmetric Property (of Equality)Example: If a = b then b = a.Hint: To be symmetric two things must be the same. In math twonumbers must be the same for it to be the symmetric property

C. Transitive Property (of Equality)Example: If a = b and b = c then a = c.Hint: Transitive starts with a “T” as well does three. There must be three number that equal the same thing to be the transitive property.

Associative Property of Adding and Multiplication

D. Associative Property of Adding

Example: (3 + 4) + 5 = 3 + (4 + 5)

E. Associative Property of Multiplication

Example: a(bc)=(ab)c

Hint: Think associative as grouping. In this property you are grouping numbers

together with parentheses.

Commutative Property of Addition and Multiplication

-This property is simply moving numbers

around in a number problem

F. Commutative Property of Addition

Example: 5 + 6 = 6 + 5

G. Commutative Property of Multiplication

Example: 5*8=8*5

Hint: You are commuting the numbers to

different places in the problem.

Distributive Property of Multiplication over Adding

I. Property of Opposites or Inverse Property of Addition   

Examples: a(b + c) = ab + ac  

(b + c)a = ba + ca

Hint: You are distributing a to b and c. It’s an easy property to grasp.

Property of Opposites or Inverse Property of Addition

I. Property of Opposites or Inverse Property of Addition

Examples: a3+(-a3)=0 6+(-6)=0

Hint: This is quite a simple property. It is basically taking the opposite of the

number either it being negative or positive to make the problem equal zero

Identity Property of Addition

J. Identity Property of Addition

Example: x+0=x, a2+0=a2

Hint: An extremely easy property to remember. Simply adding zero to the problem will make the problem an

identity property of addition.

Identity Property of Multiplication

K. Identity Property of Multiplication

Examples: x*1=x,

6*1=6

Hint: Multiplying by one is the only way an identity multiplication problem can

work

Multiplicative Property of Zero

L. Multiplicative Property of Zero

Examples:18*0=0

x*0=0

Hint: Your goal in this problem is to get zero. Anything multiplied by zero will

equal zero

Closure Property of Addition and Multiplication

M. Closure Property of Addition Explained: Closure property of real number addition states that the sum of any two real numbers equals another real number.Example: 5,7 are both real numbers

5+7=13, 13 is a real numberN. Closure Property of Multiplication

Explained: Closure property of real number multiplication states that the product of any two real numbers equals another real number.

Example: 2, 3 are both real numbers 2*3=6, 6 is a real number

Hint: Real numbers are rational numbers. A rational number is a number that can be written as a simple fraction i.e. 3= 3/1.

Zero Product Property

O. Zero Product Property

Explained: If two numbers equal zero one number in the problem must be zero

Example: ab=0, a=0 b=0

Product of Roots Property

P. Product of Roots Property

Explained: For all positive real numbers a and b

Example:

Hint: That is, the square root of theproduct is the same as the product of the square roots.

Quotient of Roots Property Q. Quotient of Roots Property

Explained:For all positive real numbers a and b, b ≠ 0

Example:

Hint: The square root of the quotient is the same as the quotient of the square roots.

Root of a Power Property

S. Root of a Power Property

Addition Property and Multiplication Property

Addition

If the same number is added to both sides of an equation, the two sides remain equal. That is, if x = y, then x + z = y + z.

MultiplicationFor all real numbers  a  and  b , and for  c ≠ 0 ,

a = b is equivalent to ac = bc .

Power of a Product Property

Explained: When a number is repeatedly multiplied by itself, we get the powers of that number

Examples: x5 = xxxxx

62=(6)(6)=36

Power of a Power Property

Explained: A power of a power is a problem that involves multiplying two powers together. An example will explain this better Examples: (32)2=34

(x7)2=x14

Hint: If you see a power within a parentheses and one outside you simply multiply the exponents

Zero Power Property

Explained: If an exponent of a number is 0 then it automatically becomes one. This does trick people into thinking it is 0, but it is really only one.

Example: 134342340=1

x0=1

Power of a QuotientPOWER OF A QUOTIENT

Explained: The law of exponents for a power of an indicated quotient may be developed from the following

Problem •

Therefore,•

• The law is stated as follows: The power of a quotient is equal to the quotient obtained when the dividend and divisor are each raised to the indicated power separately, before the division is performed.

Property of Reciprocals or Inverse Property of Multiplication

Explained: A reciprocal is the number you have to multiply a given number by to get one.

Example: 1/2x=1

1/2(2)x=1(2)

x=2

The reciprocal of 1/2 is 2 or 2/1

Hint: Another name for "reciprocal" is "multiplicative

inverse."

Power of a Root Property

Negative Power ProductExplained: A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side. For instance, "x–2" (x to the minus two) just means "x2, but underneath, as in 1/(x2)".

Examples: Write x–4 using only positive exponents. •

Write x2 / x–3 using only positive exponents. •

Write 2x–1 using only positive exponents. •  

Note that the "2" above does not move with the variable; the exponent is only on the "x". Write (3x)–2 using only positive exponents. •

Property Quiz8+7=7+38

a*b*c=c*b*a

a(b+7)= ab+ 7a

a = b then b = a

5,7 are both real numbers 5+7=13

x*0=0

6*1=6

2, 3 are both real numbers 2*3=6

x+0=x

6+(-6)=0

a = b and b = c then a = c

a+(b+c)= (a+b)+c

a(bc)=(ab)c

a. Commutative Property of Multiplication

b. Zero Property Product

c. Product of Roots

d. Symmetric Property

e. Reflexive Property

f. Transitive Property

g. Closure Property of Addition

h. Closure Property of Multiplication

i. Identity Property of Multiplication

j. Identity Property of Addition

k. Property of Opposites or Inverse Property

of Addition

l. Associative Property of Addition

m. Associative Property of Multiplication

n. Commutative Property of Addition

o. Distributive Property

p. Quotient of Roots Property

q. Product of Roots Property

r. Negative Power Property

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p

r

Solving 1st power inequalities in one variable

Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative. The easiest way to show this is with some examples:

                                                           

Solve 6x − 3 < 2x + 5

To finish this problem simply divide by four on each side. The final product should be x<2.

Solve 6x > 12

6x > 12

6 6

x > 2

<,> are greater than or equal to.

x < 0

This is a special case. This symbol means absolute value. Absolute value means the distance the number is from zero so the number can not be negative. This answer would be null set or 0

x > 0

This is another special case. Any number put into this will work meaning it is all real numbers or {all reals}

Inequalities Continued

T. Conjunctions

Explained: A conjunction is a mathematical operator that returns an output of true if and only if all of its answers are true.

Example: -2 < x < 4

Answer Graphed:

Hint: When graphing the solution set it must be connected for it to be a conjunction. If it is not then

it can not be a conjunction. Also, If it is <,> there is an open circle. If it is <,> the circle is shaded

Inequalities Continued

U. Disjunctions

Explained: A disjunction of two statements is formed by connecting them with the word "or." A disjunction is true when one or both statements are true. The solution set of a disjunction is the union of the two graphs. 

Example: -2x-6>4 or x+5>4 (You would solve this problem like the previous slides)

Answer Graphed:

Hint: Disjunctions can be graphed no matter what. If they over lap they are still disjunctions.

Linear EquationsUnderstanding the Y-intercept Form

A linear equation in this form is called the y-intercept form or slope-intercept form.In this form, you can just LOOK at the equation and pick out the important information that you need to graph the line. This is the reason why the y-intercept form is preferred over the standard form! In this format, "b" is where the line crosses the y-axis and "m" is the slope of the line.

The place where the line crosses the y-axis is called the y-intercept.General Form

Ax + By + C = 0General Form is can be put into Y-interceptform by simplifying.Example:6x+2y=4 2y=-6x+4 2= 2 y=-3x+2

Linear Equations Cont.

Use to slope formula to help complete a linear equation. The slope m of the line through the points (x1, y1) and (x

2, y 2) is given by

If you can not find the y intercept of the line use the point slope formula. Simply plug in the numbers to find it.

Linear Equations Cont.

Analyzing Slope

m=slope in y intercept form

m<0 The slope is negative making the line fall like this.

m>0 The slope is positive making the slope rise like this.

Linear Equations Cont.

How to graph a line:

Linear Equations in Two Variables

Explained: A system of linear equations is two or more linear equationsthat are being solved simultaneously. In general, a solution of a system in two variables is an ordered pair that makes BOTH

equations true.  In other words, it is where the two graphs intersect, what they have in common.  So if an ordered pair is a solution to one equation, but not the other, then it is NOT a solution to the system.

• Vocab-A consistent system is a system that has at least one solution. -An inconsistent system is a system that has no solution. -The equations of a system are dependent if ALL the solutions of one equation are also solutions of the other equation.  In other words, they end up being the same

line. -The equations of a system are independent if they do

not share ALL solutions. They can have one point in common, just not all of them.

Linear Equations in Two Variables Cont.

One Solution Explained:

If the system in two variables has one solution, it is an ordered pair that is a solution to BOTH equations.  In other words, when you plug in the values of the ordered pair it makes BOTH equations TRUE. 

If you do get one solution for your final answer, is this system consistent or inconsistent?

Answer: Consistent

If you do get one solution for your final answer, would the equations be dependent or independent?

Answer: Inconsistent

The graph to the side illustrates a system of two equations and two unknowns that has one solution:

Linear equations in Two Variables Cont.

No Solution ExplainedIf the two lines are parallel to each other, they will never intersect.  This means they do not have any points in common.  In this situation, you would have no solution.

If you get no solution for your final answer, is this system consistent or inconsistent?

Answer: Inconsistent

If you get no solution for your final answer, would the equations be dependent or independent?

Answer: Independent

The graph to the side illustrates a system

of two equations and two unknowns that

has no solution:

Hint: Perpendicular lines never have solutions

Linear equations in Two Variables Cont.

Infinite Solutions Explained:

If the two lines end up lying on top of each other, then there is an infinite number of solutions.  In this situation, they would end up being the same line, so any solution that would work in one equation is going to work in the other.

If you get an infinite number of  solutions for your final answer, is this system consistent or inconsistent?

Answer: Consistent

If you get an infinite number of  solutions for your final answer, would the equations be dependent or independent?

Answer: Dependent

The graph to the side illustrates a system

of two equations and two unknowns that

has an infinite number of solutions:

Linear SystemsSolving By Substitution Explained:

The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works.

2x – 3y = –2

4x +  y = 24

y = –4x + 24 4x + y = 24

Here you are isolating “y” so you can substitute it for “y” in the other system.

Now simply plug “y” in for x and solve.

2x – 3(–4x + 24) = –22x + 12x – 72 = –214x = 70x = 5

Now you can plug this x-value back into either equation, and solve for y. But since there already exists an expression for "y =", it will be simplest to just plug into this:y = –4(5) + 24 = –20 + 24 = 4Then the solution is (x, y) = (5, 4).

Linear Systems Cont. Solving by Addition / Elimination Explained

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations. If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.

Solve the following system using addition. 2x + y = 93x – y = 16

Hint: Note that the y's will cancel out. So draw an "equals" bar under the system, and add the like terms:2x + y = 93x – y = 165x      = 25

Now you can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so we will back-solve in that one:2(5) + y = 9  10 + y = 9          y = –1

Then the solution is (x, y) = (5, –1).It doesn't matter which equation you use for the back solving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:3(5) – y = 16  15 – y = 16        –y = 1          y = –1

FactoringFactoring out the Greatest Common Factor (GCF) is perhaps the most used type of factoring because it occurs as part of the process of factoring other types of products. Before you can factor trinomials, for example, you should check for any GCF.

#1: Factor the following problem completely 2x-14

Look for the greatest factor common to every term

Answer: 2

Factor out the GCF by dividing it into each term

Answer 2(x-7)

Oftentimes when there is no factor common to all terms of a polynomial there will be factors common to some of the terms. A second technique of factoring called grouping is illustrated in the following examples.

#2. Factor the following problem completely 3ax+6ay+4x+8y

Factor out 3a from the first 2 terms and 4 from the last 2 terms.

Answer:

Notice that the terms inside each set of parentheses are the same. Those terms have now become the GCF. The answer may be checked by multiplying the factored form back out to see if you get the original polynomial.

Final Answer: (3a+4)(x+2y)

Hint: Grouping is only effective if there is a GCF between factors like in this problem (x+2y)

Factoring Cont.A difference in two perfect squares by definition states that there must be two terms, the sign

between the two terms is a minus sign, and each of the two terms contain perfect squares. The answer after factoring the difference in two squares includes two binomials. One of the binomials contains the sum of two terms and the other contains the difference of two terms. In general, we say

#6: Factor the following problem completely a. Examine the problem for a GCF. There is none. 2. To factor a difference in two squares, use two sets of parentheses. 3. Take the square root of each term. The square root of a variable’s exponent will be half of the

exponent. and

4. Use the square roots to fill in the parentheses. Be sure to check that neither factor will factor again. What is the final answer?

Final Answer: (3x+4y)(3x-4y)

Factoring Cont.Factoring the sum or difference in two perfect cubes is our next technique. As with squares, the difference in two cubes

means that there will be two terms and each will contain perfect cubes and the sign between the two terms will be negative. The sum of two cubes would, of course, contain a plus sign between the two perfect cube terms. The follow formulas are helpful for factoring cubes: Sum: Difference: Notice that the sum and the difference are exactly the same except for the signs in the factors. Many students have found the acronym SOAP extremely helpful for remembering the arrangement of the signs. S represents the fact that the sign between the two terms in the binomial portion of the answer will always be the same as the sign in the given problem.

O implies that the sign between the first two terms of the trinomial portion of the answer will be the opposite of the sign in the problem.

AP states that the sign between the final two terms in the trinomial will be always positive. Factor the following problem completely This is a difference in two cubes, so begin with two sets of parentheses. • In the first set, there will be a binomial containing the cube root of each term. In this problem, x and 3. • In the second set there will be a trinomial. The first term of the trinomial is the square of the first term in the binomial. • The last term is the square of the last term in the binomial. • The middle term is the product of the two terms in the binomial. • You will be finished when you insert the appropriate sign between each of the terms. • (x-3)(x2+3x+9)

Factoring Cont.Before factoring a trinomial, examine the trinomial to be sure that terms are

arranged in descending order. Most of the time trinomials factor to two binomials in product form.

Factor the following problem completely. The three terms are arranged in descending order. There is not a GCF.

Therefore the factoring process is begun by opening two sets of parentheses. • Place the factors for the first term of the trinomial in the front of each set of

parentheses. • Then, because the sign of the last term is positive, factor the last term of the

trinomial to factors that multiply to give 12 and add to give 7. • Finally, because the sign of the last term is positive, the sign of the 4 and the

sign of the 3 will each have the same sign. Because the sign of the 7 ispositive, the sign of the 4 and the sign of the 3 will each be a positive sign.Check the answer using multiplication.

• (x+4)(x+3)

Factoring Cont.A general trinomial is one whose first term has a coefficient that can not be factored out as a GCF. The method

of trial and error will be used to mentally determine the factors that satisfy the trinomial. We will show you the steps to factor each of the following general trinomials completely.

Factor the following problem completely.

Factor out the GCF. •

In factoring the general trinomial, begin with the factors of 12. These include the following: 1, 12, 2, 6, 3, 4. As a general rule, the set of factors closest together on a number line should be tried first as possible factors for the trinomial.

•

The only factors of the last term of the trinomial are 1 and 3, so there are not other choices to try. Because the last term is negative the signs of the factors 1 and 3 must be opposite.

•

This is the first trial.  The answer must be checked by multiplication, as follows: •

Rational Expressions

Rational Expressions Cont.

Solve Problems

First Factor

Once Factored simply cancel like terms

Final Answer

Rational Expressions Cont.Addition and Subtraction of Rational Functions

Explained: To add and subtract rational functions, we follow the same method as fractions.  

Step 1  Factor everything and find the least common denominator.

Step 2 Multiply the numerators and the denominators by the appropriate denominator so that the denominator becomes the least common Denominator.

Step 3  Add the numerators together.

Step 4  Factor the numerator.

Step 5 Cancel any common factors.

Multiplication of Rational Functions

Recall that when we multiply fractions we first cross cancel. When we multiply rational expressions we follow the same approach.  First we factor then we cross cancel.  

 x2 - 2x + 1      x2 + 4x + 3        First Factor      x  + 1               x - 1

(x - 1)2       (x + 3)(x + 1)           Cancel the x + 1 and the x - 1 and the x - 1     x + 1            x - 1 =   (x - 1)(x + 3)

Rational Functions Cont.

Division of Rational Functions

Explained: Essentially division of rational functions is the same as multiplication. Instead you flip the term behind the division side.

Example:

Quadratic Equations in One Variable

Quadratic Equations in One VariableExplained: A quadratic equation in x is any equation that may be written in the formax2 + bx + c = 0, where a, b, and c are coefficients and a ≠ 0.Note that if a=0, then the equation would simply be a linear equation, not quadratic.

Examplesx2 + 2x = 4 is a quadratic since it may be rewritten in the form ax2 + bx + c = 0 by applying the Addition Property of Equality and subtracting 4 from both sides of =.(2 + x)(3 – x) = 0 is a quadratic since it may be rewritten in the form ax2 + bx + c = 0 by applying the Distributive Property to multiply out all terms and then combining like terms. x2 - 3 = 0 is a quadratic since it has the form ax2 + bx + c = 0 with b=0 in this case.

Solving Quadratic Equations – Method 1 - FactoringThe easiest way to solve a quadratic equation is to solve by factoring, if possible.Here are the steps to solve a quadratic by factoring:1. Write your equation in the form ax2 + bx + c = 0 by applying the DistributiveProperty, Combine Like Terms, and apply the Addition Property of Equality tomove terms to one side of =.2. Factor your equation by using the Distributive Property and the appropriatefactoring technique. Note: Any type of factoring relies on the DistributiveProperty.3. Let each factor = 0 and solve. This is possible because of the Zero Product

Law.Example: Solve (3x + 4)x = 7(3x + 4)x = 7 Given3x2 + 4x = 7 by the Distributive Property3x2 + 4x – 7 = 0 by the Addition Property of EqualityNow, factor 3x2 + 4x – 7 = 0This factors as (3x + ?)(x - ?) = 0 or (3x - ?)(x + ?) = 0 where the two unknownnumbers multiply to -7 when we use the Distributive Property to multiply out.Also the first two terms must multiply out to 3x2. The middle products must addup to 4x.(3x + 7)(x - 1) = 0 gives us middle products 7x and –3x adding up to 4x.

Solving Quadratic Equations – Method 2 – Extracting Square RootsExtracting square roots is a very easy way to solve quadratics, provided the equation is in the correct form. Basically, Extracting Square Roots allows you to rewrite x2 = k as x = ±√k, where k is some real number. Algebraically, we are taking square roots of both sides of the equation as shown below and inserting the ± to account for both a positive and negativecase. Note that the squared quantity must be isolated on one side of = before you can extract the square roots.

Example: Solve x2 = 9 by extracting square roots

Example: Solve (2x – 5)2 + 5 = 3(2x – 5)2 + 5 = 3 Given(2x – 5)2 = -2 Addition Property of Equality used to add –5 to both sides√ (2x – 5)2 = ±√(-2) Extract Square Roots2x – 5 = ± i√2 Simplify Radicals and Apply Definition of “i”2x = 5 ± i√2 Addition Property of Equalityx = (5 ± i√2) / 2 Division Property of Equality

Solving Quadratic Equations – Method 3 – Completing The SquareThis method of solving quadratic equations is straightforward, but requires a specific sequence of steps. Here is the procedure:Example: Solve 3x2 + 4x – 7 = 0 By Completing The Square1. Isolate the x2 and x-terms on one side of = by applying the Addition Property of Equality.3x2 + 4x = 72. Apply the Division Property of Equality to divide all terms on both sides by the coefficient on x2.(3x2)/3 + (4x)/3 = 7/3x2 + (4/3)x = 7/3(3x2)/3 + (4x)/3 = 7/3x2 + (4/3)x = 7/3 Note: Steps 1 and 2 may be done in either order.3. Take ½ of the coefficient on x. Square this product. Add this square to both sides using the Addition Property of Equality. In this case, we take ½ of 4/3 which is (1/2)•(4/3) = 4/6. Square 4/6 to get (4/6) •(4/6) = 16/36 = 4/9 when reduced. Add 4/9 to both sides to getx2 + (4/3)x + 4/9 = 7/3 + 4/9x2 + (4/3)x + 4/9 = 21/9 + 4/9 multiply 7/3 by 3/3 to get common denominatorx2 + (4/3)x + 4/9 = 25/9 add fractions4. Factor the left side. Note: It will always factor as (x ± the square root of what you added)2(x + 2/3)2 = 25/95. Solve by extracting square roots.√ (x + 2/3)2 = ±√(25/9) Extract Square Rootsx + 2/3 = ±5/3 Simplify Radicalsx = -2/3 ± 5/3 Addition Property of Equality

Solving Quadratic Equations – Method 4 – Using The Quadratic FormulaSolving a quadratic equation that is in the form ax2 + bx + c = 0 only involves plugging a, b, and c into the formula

Example: Solve (x + 3)2 = x – 2(x + 3)2 = x – 2 Given(x + 3)(x + 3) = x – 2 Rewritex2 + 6x + 9 = x – 2 Multiply out with Distributive Property, Combine Like Termsx2 + 5x + 11 = 0 Addition Property of Equality - add 2, add –x to both sidesPlug a=1, b=5, c =11 from 1x2 + 5x + 11 = 0 into the Quadratic Formula to get

which simplifies to

after we simplify the radical and rewrite √(-19) as (√19) • i by applying thedefinition of i.

Quadratic DiscriminantThe discriminant is a number that can be calculated from any quadratic equation A quadratic equation is an equation that can be written as

ax ² + bx + c where a ≠ 0

The discriminant in a quadratic equation is found by the following formula and the discriminant provides critical information regarding the nature of the roots/solutions of any quadratic equation.

discriminant= b² − 4ac

Positive Discriminant b² − 4ac > 0 Two Real Solutions If the discriminant is a perfect square the roots are rational. Otherwise, they are irrational.

Discriminant of Zero

b2-4ac=0

There will be one solution

Negative Discriminant

b2-4ac<0

There are no real solutions

Functions

F of xf(x) means the function of x. It's a short hand way of saying the values that make up a line which are found from some function (or equation) based on x.

Are all relations functions?Functions are relations only when every input has a distinct output, so no, not all relations are functions but all functions are relations.

State the domain and range of the following relation. Is the relation a function?{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}The above list of points, being a relationship between certain x's and certain y's, is a relation. The domain is all the x-values, and the range is all the y-values. To give the domain and the range, I just list the values without duplication:domain:  {2, 3, 4, 6}ange:  {–3, –1, 3, 6}

Functions Cont.Given to order pairs of data, find a linear function that contains those points.

This question can be easily done in a few steps

1. Find the Slope with the slope formula

2. Use the slope-point formula to find y intercept

These steps would look like this in a problem

{(1,2)(2,6)}

1. 6-2 = 4 M=4

2-1 = 1

2. y-2=4(x-1)

=y-2=4x-2

=y=4x

Here you are using the slope formula

Here you are using the slope-point formula

to find the y intercept.

Graphing ParabolasGraphing the Parabola y = ax2 + bx + c1.    Determine whether the parabola opens upward or downward.

a.    If a > 0, it opens upward.b.    If a < 0, it opens downward.

2.    Determine the vertex.a.    The x-coordinate is .b.    The y-coordinate is found by substituting the x-coordinate, from step 2a, in the equation y = ax2 + bx + c.

3.    Determine the y-intercept by setting x = 0.4.    Determine the x-intercepts (if any) by setting y = 0, i.e., solving the

equation        ax2 + bx + c = 0.

5.    Determine two or three other points if there are no x-intercepts.

Graphing Parabolas Cont.

Graph of y = x 2

Simplifying Expressions with Radicals

When presented with a problem like , we don’t have too much difficulty saying that the answer 2 (since ). Even a problem like is easy once we realize . Our trouble usually occurs when we either can’t easily see the answer or if the number under our radical sign is not a perfect square or a perfect cube.

A problem like may look difficult because there are no two numbers that multiply together to give 24. However, the problem can be simplified. So even though 24 is not a perfect square, it can be broken down into smaller pieces where one of those pieces might be perfect square. So now we have .

Simplifying a radical expression can also involve variables as well as numbers. Just as you were able to break down a number into its smaller pieces, you can do the same with variables. When the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, etc). When the radical is a cube root, you should try to have terms raised to a power of three (3, 6, 9, 12, etc.). For example,

These types of simplifications with variables will be helpful when doing operations with radical expressions. Let's apply these rule to simplifying the

Simplifying Expressions with Radicals

Examples:

Line of Best Fit or Regression Line

A regression line is a line drawn through a scatter plot of two variables. The line is chosen so that it comes as close to the points as possible. A graphing calculator is very helpful for finding this. Below is a scatter plot with a regression line in it.

Word Problems

If Sally can paint a house in 4 hours, and John can paint the same house in 6 hours, how long will it take for both of them to paint the house together?

a. 2 hours and 24 minutes

b. 3 hours and 12 minutes

c. 3 hours and 44 minutes

d. 4 hours and 33 minutes

SolutionIf Sally can paint the house in 4 hours, then in 1 hour she can paint 1/4 of the house. If John can paint the house in 6 hours, then in 1 hour he can paint 1/6 of the house. Let x = number of hours it would take them together to paint the house, then working together, in 1 hour they can paint 1/x of the house.

The equation is this:

Multiply both sides by the LCD which is 12x:

= 2.4 hours or A

Word Problem

If 2 pens and 3 notebook cost 4.55 and 3 pens and 2 note books cost 3.70, find the price of each pen

a. 1.25

b. .50

c. .40

d. 1.00

e. .67

Solution

P: Pens

N: Notebooks

-2(2p+3n=4.55)

3(3p+2n=3.70) Use elimination method.

p=.40 or C

Word Problem

The Sales price of a car is 12,590, which is 20% off the original price. What is the original price?

a. 14,000

b. 14,670.30

c. 17,894

d. 15,737.50

Solution

Multiply .2 by 12,950 and then add that number to 12,950 to get 15,737.50 or D

Word Problem

Employees of an appliance store receive an additional 20% off of the lowest price on an item. If an employee purchases a dishwasher 15% off sale, how much will he pay if the dish washer orginally cost 450?

a. 287.96b. 333.39c. 306

Solution

Simply find how much 15% of 450 is subtract that from 450. Take that subtracted number and multiply it by 20% and subtract that from that number to get 306

Citations • http://www.tutorvista.com/content/math/algebra/linear-two-variable/linear-equations-

two-variable.php • http://hotmath.com/search/hotmath-search.jsp?term=disjunction • http://www.chacha.com/topic/properties-of-square-roots • http://philosophy.lander.edu/logic/conjunct.html • http://rachel5nj.tripod.com/NOTC/ssoewog2.html • http://www.tutorvista.com/content/math/algebra/linear-two-variable/linear-equations-

two-variable.php• http://honolulu.hawaii.edu/distance/sci122/SciLab/L6/bestfit.html• http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_radical_simplify.xml• http://a-s.clayton.edu/garrison/math%200099/parabola.htm• http://www.purplemath.com/modules/rtnldefs.htm• http://www.mathnstuff.com/math/spoken/here/2class/320/quadequ.htm• http://www.tpub.com/math1/17.htm• http://www.sosmath.com/diffeq/system/linear/basicdef/basicdef.html

FIN