quadratic equations solving quadratic equations. 7/9/2013 quadratic equations 2 solving quadratic...

Quadratic Equations

Solving

Quadratic Equations

Quadratic Equations 27/9/2013

Solving Quadratic Equations

Standard Form

ax2 + bx + c = 0

with a ≠ 0

Solve to Find: Solutions and Solution Set

What is a solution ? What is a solution set ?

How many solutions ?


Solving Quadratic Equations Standard Form

ax2 + bx + c = 0

with a ≠ 0 Solve by:

Square Root Property Completing the Square Quadratic Formula Factoring


Solving Quadratic Equations Solutions

A solution for the equation

ax2 + bx + c = 0

is a value of variable x satisfying the equation

NOTE: For now we assume all values are real numbers

The solution set for the equation is the set of all of its solutions


Solving Quadratic Equations Examples

1. x2 + 2x – 3 = 0

2. 16x2 – 8x + 1 = 0

3. x2 + 2x + 5 = 0

Solutions: –3, 1 Solution set: {–3 , 1}

Solution: 14

Solutions: noneSolution set: { } OR

Solution set: { }14


The Square Root Property Reduced Standard Form - Type 1

ax2 = c , for a ≠ 0

Rewrite:

Square Root Property

If x2 = k then

OR

cax2 = = k

x2 = x│ │

k =x

k =x –


The Square Root Property Square Root Property

If x2 = k then

ORx2 = x│ │

k =x

k =x –Note:

This does NOT say that This DOES say that IF x2 = 4 then

OR

The Principal Root The Negative Root

= 2x = 4

2 4 =

4 –x = = –2


Square Root Property Roof repairs on a 100-story building

A tar bucket slips and falls to the sidewalk

If each story is 12.96 feet, how long does it take the bucket to hit the sidewalk?

Distance of fall: (12.96 ft/story)(100 stories) = 1296 ft

For a fall of s ft in t seconds we have

s = 16t2 OR t2 = s/16


Square Root Property Roof repairs on a 100-story building

Solving for t

We choose the positive value of t Thus

s = 16t2 OR t2 = s/16

OR =t s 4

–=t4 s

WHY ?

436

= 9= seconds=t4 s

=4

1296


Special Forms Perfect-Square Trinomials

Always the square of a binomial

A2 + 2AB + B2 = (A + B)2

A2 – 2AB + B2 = (A – B)2 … OR (B – A)2


Special Forms Examples

x2 + 6x + 9

4y2 – 20y + 25

9y2 + 12xy + 4x2

x2y2 – 6xyz + 9z2

= (x + 3)2

= (2y – 5)2

= (3y + 2x)2

= (xy – 3z)2


Completing the Square Convert to perfect-square trinomial

Solve using square root property

Example 1:

2x2 – 3x + 5 = x2 – 9x + 12

x2 + 6x = 7

To make the left side

we need to find a such that

(x + a)2 = x2 + 2ax + a2

Subtract x2 and 5, on each side, and add 9x

= x2 + 6x + a2

a perfect square



Example 1:

2x2 – 3x + 5 = x2 – 9x + 12

x2 + 6x = 7

(x + a)2 = x2 + 2ax + a2 = x2 + 6x + a2

So 2ax = 6x giving a = 3 and a2 = 9Adding 9 to each side of x2 + 6x = 7x2 + 6x + 9 = 7 + 9 = 16

(x + 3)2 = 16 x + 3 16 = + = +4

x = –3 + 4 Solution set: { –7, 1 }



Example 2:

5y2 – 87y + 89 = 25 + 15y – 4y2

9y2 – 102y = –64

For the left need to find a such that

(3y + a)2 = 9y2 + 6ay + a2

= 9y2 – 102y + a2

Add 4y2 to each side, subtract 89 and 15y

to be a perfect square we


Completing the Square Example 2:

5y2 – 87y + 89 = 25 + 15y – 4y2

9y2 – 102y = –64

(3y + a)2 = 9y2 + 6ay + a2 = 9y2 – 102y + a2

So 6ay = –102y giving a = –17 and a2 = 289

Adding 289 to each side

9y2 – 102y + 289 = –64 + 289Thus

(3y – 17)2 = 225


Completing the Square Example 2:

5y2 – 87y + 89 = 25 + 15y – 4y2

9y2 – 102y = –64

(3y – 17)2 = 225

3y – 17 225 = +15 = +

3y

= 17 15 +

y

= +17 3

15 3 =

32 3

2 3

Solution Set: { 32 3

2 3

, }


The Quadratic Formula I Complete the Square on the Standard Form

ax2 + bx + c = 0 , for a ≠ 0 Rewrite: move c, divide by a

ca

ba –x2 x + =

Now complete the square on the left side Want a constant k such that

bax2 x + + k2

is a perfect square – that is bax2 x + + k2 = x + k( )2


The Quadratic Formula II Complete the Square on the Standard Form

ca

ba –x2 x + =

bax2 x + + k2 = x + k( )2

= x2 + 2xk + k2

Thus 2k = ba … and = b

2a( )2k2

bax2 x + c

a–= + b2a( )2

+ b2a( )2

x

+ b2a( )2 c

a–= + b4a

2

2


The Quadratic Formula III Complete the Square on the Standard Form

x +( b

2a)2

= ca– + b2

4a2

+ b2

4a2= – 4ac4a2 =

4a2b2 – 4ac

x + b2a =

4a2b2 – 4ac+

x =b

2a 2ab2 – 4ac+

x = 2ab b2 – 4ac+

Square Root Property

Quadratic Formula


Quadratic Formula Examples Example 1

Solve: 7x2 + 9x + 29 = 27

First put into standard form:

7x2 + 9x + 2 = 0

x =2ab2 – 4ac ±–b Apply Quadratic

Formula

Thus a = 7 , b = 9 , c = 2

WHY ?

x = ±–9 92 – 4 7 2 ( ( ))

2 7( )



Solve: 7x2 + 9x + 29 = 27

x = ±–9 92 – 4 7 2 ( ( ))

2 7( )

=±–9 81 – 56

14 =

±–9 25 14

=±–9 5

14

Question: Are there always two solutions ?

Solution set:

{–1, } 72 –

=

–14 14

(continued)

– 4 14



Solve: 4x2 – 12x + 29 = 20

First put into standard form:

4x2 – 12x + 9 = 0 WHY ?

Thus a = 4 , b = –12 , c = 9

x =2ab2 – 4ac ±–b Apply Quadratic

Formula

x =2 4( )

–12 2 – 4 4 9 ±12 ( ( ))( )



Solve: 4x2 – 12x + 29 = 20

Solution set:

x =2 4( )

–12 2 – 4 4 9 ±12 ( ( ))( )

(continued)

x =8

144 – 144±12

= 812 x

= 23 x {

23 }

Question: Are there always two solutions ?


The Quadratic Formula The Discriminant

For ax2 + bx + c = 0 we found that

The expression b2 – 4ac is called the discriminant

It determines the number and type of solutions

x =2ab2 – 4ac ±–b


The Quadratic Formula The Discriminant

Determines the number and type of solutions

If b2 – 4ac = 0 have one real solution If b2 – 4ac > 0 have two real solutions If b2 – 4ac < 0 have two complex

solutions


Factoring Reduced Standard Form – Type 2

ax2 + bx = 0 , for a ≠ 0 Factor x and use zero-product property

x(ax + b) = 0 Zero-product Property:

if and only if p = 0 or q = 0 ... or both

For any real numbers p and q , pq = 0


Factoring Reduced Standard Form – Type 2

ax2 + bx = 0 , for a ≠ 0 Example:

Solve 3x2 – 5x = 0

x(3x – 5) = 0

From the zero-product property

x = 0 or 3x – 5 = 0 or both

Solution Set: {0 , 53 }


Special Forms Conjugate Binomials

Pairs of binomials of form (A + B) and (A – B) are called conjugate

binomials

Product of conjugate pair is always a difference of squares

(A + B)(A – B) = A2 – B2


Special Forms Conjugate Binomials

Examples

x2 – 4

9 – 4x2

4x2 – 9y2

= (x + 2)(x – 2)

= (3 – 2x)(3 + 2x)

= (2x – 3y)(2x + 3y)


Not-So-Special Forms What if the expression is not a

special form? We look for factors with integer

coefficients

In general these have form:

x2 + (a + b)x + ab = (x + a)(x + b)

We look for factors (x + a) and (x + b) that fit this form


Not-So-Special Trinomials What if the expression is not a

special form?

Factoring Examples:

x2 + 7x + 12 (x + 3)(x + 4)=

y2 + 13y + 40 (y + 5)(y + 8)=

(2x + 3)(2x + 5)=4x2 + 16x + 15

Question: Is there a systematic way to do this ?


Not-So-Special Trinomials Factoring Trinomials with negative

constant term In general these have form:

x2 + (a – b)x – ab = (x + a)(x – b)

We look for factors that fit this form


Not-So-Special Trinomials Examples

x2 – x – 12 =

y2 + 3y – 40 =

4x2 – 4x – 15 =

Note the sign of last term: always negative

Note the sign of the middle term: positive or negative

(x + 3)(x – 4)

(y – 5)(y + 8)

(2x + 3)(2x – 5)

WHY ?

WHY ?


Factor x2 – 5x – 24

x2 + (a + b)x + ab = (x + a)(x + b)

Trinomials with two negative terms

Not-So-Special Trinomials

Terms a , b of opposite sign

Sign of the larger term

Find a and b such that

Possible factors of –24 :

1, –24 2, –12 3, –8 4, –6

Since 3 – 8 = –5 , then a = 3 and b = –8


Factor x2 – 5x – 24

Trinomials with two negative terms


Since 3 – 8 = –5 , then a = 3 and b = –8

This gives us:

NOTE: There is no solution to talk about here … since we are changing the form of an expression, NOT solving an equation

x2 – 5x – 24 = (x + 3)(x – 8)


Factor x2 + 5x – 24

x2 + (a + b)x + ab = (x + a)(x + b)

Trinomials with one negative term– the constant


Terms a , b of opposite sign

Sign of the larger term


Possible factors of –24 :

–1, 24 –2, 12 –3, 8 –4, 6

Since 3 – 8 = 5 , then a = 8 and b = –3


Factor x2 + 5x – 24

Trinomials with one negative term– the constant


Since 3 – 8 = 5 , then a = 8 and b = –3

This gives us:x2 + 5x – 24 = (x + 8)(x – 3)

NOTE: This is not a solution , since we arechanging the form of an expression , NOT solving an equation


Solve x2 – 8x + 12 = 0 by factoring

x2 + (a + b)x + ab = (x + a)(x + b)

Trinomials with one negative term– the middle term


Terms a , b of same sign

Sign of both terms


Possible factors of 12 :

–1, –12 –2, –6 –3, –4

Since –2 – 6 = –8 , then a = –2 and b = –6


Solve x2 – 8x + 12 = 0 by factoring

Trinomials with one negative term– the middle term


Since –2 – 6 = –8 , then a = –2 and b = –6

x2 – 8x + 12 = (x – 2)(x – 6) = 0 This gives us:

NOTE: This is a solution , since we aresolving an equation

So x – 2 = 0 or x – 6 = 0

Solution Set: 2, 6{ }


Solve x2 + 11x + 28 = 0 by factoring

x2 + (a + b)x + ab = (x + a)(x + b)

Trinomials with no negative terms


Terms a , b of same sign

Sign of both terms


Possible factors of 28 :

1, 28 2, 14 4, 7

Since 4 + 7 = 11 , then a = 4 and b = 7


Solve x2 + 11x + 28 = 0 by factoring

Trinomials with no negative terms


Since 4 + 7 = 11 , then a = 4 and b = 7

x2 + 11x + 28 = (x + 4)(x + 7) = 0This gives us:

So x + 4 = 0 or x + 7 = 0

Solution Set: – 4 , –7{ }

NOTE: This is a solution , since we aresolving an equation


Expanding a Room

A 9 by 12 room is to be expanded by the same amount x in length and width to double the area – what is x ?

9 + x

12 ft

9 ft A1 = 108

A4 = x2A3 = 12x

A2 = 9x

x ft

x ft

12 + xOriginal area:A1 = 208 ft2

New area:A = 2A2 = 216 ft2

Two methods:

A = A1 + A2 + A3 + A4

A = (9 + x)(12 + x)= 216

= 216


Expanding a Room

Two methods:A = A1 + A2 + A3 + A4

A = (9 + x)(12 + x) = 216= 216OR

Either way: x2 + 21x + 108 = 216So x2 + 21x – 108 = 0

≈ 4.3 sq. ft.

x =2 1( )

( ( ))( )± 21 2 – 4 1 –108

21 –

x =2

± 876 21 –


Think about it !

quadratic equations solving quadratic equations. 7/9/2013 quadratic equations 2 solving quadratic...

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