Transcript Introduction to Algebra

Quadratic Equations
Solving
Quadratic Equations
Solving Quadratic Equations
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
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 ?
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Solving Quadratic Equations
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
Standard Form
ax2 + bx + c = 0
with a ≠ 0

Solve by:
 Square Root Property
 Completing the Square
 Quadratic Formula
 Factoring
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Solving Quadratic Equations

Solutions

A solution for the equation
ax2 + bx + c = 0
is a value of variable x satisfying the
equation
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
NOTE: For now we assume all values
are real numbers

The solution set for the equation is the
set of all of its solutions
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Solving Quadratic Equations

Examples

1. x2 + 2x – 3 = 0
Solutions: –3, 1 Solution set: {–3 , 1}

2. 16x2 – 8x + 1 = 0
Solution: 1
4

Solution set:
1
4
{ }
3. x2 + 2x + 5 = 0
Solutions: none
Solution set: { } OR 
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The Square Root Property

Reduced Standard Form - Type 1
ax2 = c , for a ≠ 0


Rewrite: x2 = ac = k
Square Root Property

If x2 = k then
x =k
 x2 = │x │
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OR
x = – k
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The Square Root Property

Square Root Property

If x2 = k then
x =k
 x2 = │x │
OR
x = – k
Note:
This does NOT say that  4 =  2
This DOES say that IF x2 = 4 then
x=4 =2
OR
The Principal Root
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Quadratic Equations
x = –  4 = –2
The Negative Root
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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
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Square Root Property

Roof repairs on a 100-story building
s = 16t2 OR t2 = s/16



Solving for t
s

t=
4
We choose the positive value of t WHY ?
Thus
36
1296
s


t=
= 9 seconds
=
=
4
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s

–
OR t =
4
4
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Special Forms

Perfect-Square Trinomials

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Always the square of a binomial

A2 + 2AB + B2 = (A + B)2

A2 – 2AB + B2 = (A – B)2 … OR (B – A)2
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Special Forms

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Examples

x2 + 6x + 9 = (x + 3)2

4y2 – 20y + 25 = (2y – 5)2

9y2 + 12xy + 4x2 = (3y + 2x)2

x2y2 – 6xyz + 9z2 = (xy – 3z)2
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Completing the Square

Convert to perfect-square trinomial


Solve using square root property
Example 1:
Subtract x2 and 5,
on each side, and
2x2 – 3x + 5 = x2 – 9x + 12
add 9x
2
x + 6x = 7
To make the left side a perfect square
we need to find a such that
(x + a)2 = x2 + 2ax + a2 = x2 + 6x + a2
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Completing the Square

Convert to perfect-square trinomial

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 = 9
x2
+ 6x + 9 = 7 + 9 = 16
Adding 9 to each
side of x2 + 6x = 7
(x + 3)2 = 16
x + 3 = +  16 = + 4
x = –3 + 4
Solution set: { –7, 1 }
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Completing the Square

Convert to perfect-square trinomial

Example 2:
5y2 – 87y + 89 = 25 + 15y – 4y2
9y2 – 102y = –64
Add 4y2 to each
side, subtract 89
and 15y
For the left to be a perfect square we
need to find a such that
(3y + a)2 = 9y2 + 6ay + a2
= 9y2 – 102y + a2
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Completing the Square

Example 2:
5y2 – 87y + 89 = 25 + 15y – 4y2
9y2 – 102y = –64
Adding 289
to each side
(3y + a)2 = 9y2 + 6ay + a2
= 9y2 – 102y + a2
So 6ay = –102y giving a = –17 and a2 = 289
9y2 – 102y + 289 = –64 + 289
Thus
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(3y – 17)2 = 225
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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
32
17
15
y = 3 + 3 =
Solution Set:
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{ 23 , 323 }
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2
3
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The Quadratic Formula I

Complete the Square on the Standard Form
ax2 + bx + c = 0 , for a ≠ 0
 Rewrite: move c, divide by a
x2 + ab x = – ac
 Now complete the square on the left side
 Want a constant k such that
x2 + ab x + k2
is a perfect square – that is
x2 + ab x + k2 = ( x + k)2
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The Quadratic Formula II

Complete the Square on the Standard Form
x2 + ab x = – ac
x2 + ab x + k2 = ( x + k)2
= x2 + 2xk + k2
b
Thus 2k = a … and k2 = 2ba
2
b
b
2
x + a x + 2a = – ac + 2ba
2
2
b
c
b
x + 2a = – a + 2
4a
( )
)
(
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( )
2
( )
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The Quadratic Formula III

Complete the Square on the Standard Form
(
x + 2ba
2
)
2
c
b
= –a +
4a 2
2 – 4ac
2
b
b
4
ac
= – 2 +
=
2
4a 2
4a
4a
Square Root
b
x
+ 2a =
Property
+

b 2 – 4ac
4a 2
2 – 4ac
b
b

x = 2a +
2a
Quadratic
Formula
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x =
b +  b 2 – 4ac
2a
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Quadratic Formula Examples

Example 1

Solve: 7x2 + 9x + 29 = 27
First put into standard form:
WHY ?
7x2 + 9x + 2 = 0
Thus a = 7 , b = 9 , c = 2
–b ±  b2 – 4ac
x=
2a
Apply Quadratic
Formula
–9 ±  92 – 4(7)( 2)
x=
2(7)
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Quadratic Formula Examples

Example 1 (continued)

Solve: 7x2 + 9x + 29 = 27
–9 ±  92 – 4(7)( 2)
x=
2(7)
–9 ±  81 – 56
=
14
Solution set:
2
–
–1, 7
{
}
–9 ±  25
=
14
–9 ± 5
=
=
14
–14
14
–4
14
Question: Are there always two solutions ?
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Quadratic Formula Examples

Example 2

Solve: 4x2 – 12x + 29 = 20
First put into standard form:
WHY ?
4x2 – 12x + 9 = 0
Thus a = 4 , b = –12 , c = 9
–b ±  b2 – 4ac
x=
2a
Apply Quadratic
Formula
12 ±  (–12)2 – 4 (4)(9)
x=
2(4)
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Quadratic Formula Examples

Example 2 (continued)

Solve: 4x2 – 12x + 29 = 20
12 ±  (–12)2 – 4 (4)(9)
x=
2(4)
12 ±  144 – 144
x=
8
x = 12
8
x= 3
2
Solution set:
{ }
3
2
Question: Are there always two solutions ?
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The Quadratic Formula

The Discriminant
2
 For ax + bx + c = 0 we found that
–b ±  b2 – 4ac
x=
2a


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The expression b2 – 4ac is called the
discriminant
It determines the number and type of
solutions
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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

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If b2 – 4ac < 0 have two complex
solutions
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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:
For any real numbers p and q ,
pq = 0
if and only if p = 0 or q = 0 ... or both
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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:
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{
5
0, 3
}
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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
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Special Forms

Conjugate Binomials

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Examples

x2 – 4 = (x + 2)(x – 2)

9 – 4x2 = (3 – 2x)(3 + 2x)

4x2 – 9y2 = (2x – 3y)(2x + 3y)
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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)
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We look for factors (x + a) and (x + b)
that fit this form
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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)

4x2 + 16x + 15 = (2x + 3)(2x + 5)
Question:
Is there a systematic way to do this ?
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Not-So-Special Trinomials

Factoring Trinomials with negative
constant term

In general these have form:
x2 + (a – b)x – ab = (x + a)(x – b)

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We look for factors that fit this form
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Not-So-Special Trinomials

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Examples

x2 – x – 12 = (x + 3)(x – 4)

y2 + 3y – 40 = (y – 5)(y + 8)

4x2 – 4x – 15 = (2x + 3)(2x – 5)

Note the sign of last term:
always negative
WHY ?

Note the sign of the middle term:
positive or negative
WHY ?
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Not-So-Special Trinomials

Trinomials with two negative terms
Sign of the
larger term
Terms a , b of
opposite sign
 Factor x2 – 5x – 24
 Find a and b such that
x2 + (a + b)x + ab = (x + a)(x + b)
Possible factors of –24 :
1, –24 2, –12 3, –8 4, –6
Since 3 – 8 = –5 , then a = 3 and b = –8
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Not-So-Special Trinomials

Trinomials with two negative terms
 Factor x2 – 5x – 24
Since 3 – 8 = –5 , then a = 3 and b = –8
This gives us:
x2 – 5x – 24 = (x + 3)(x – 8)
NOTE:
There is no solution to talk about here
… since we are changing the form of an
expression, NOT solving an equation
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Not-So-Special Trinomials

Trinomials with one negative term
– the constant
Sign of the
larger term
Terms a , b of
opposite sign
 Factor x2 + 5x – 24
 Find a and b such that
x2 + (a + b)x + ab = (x + a)(x + b)
Possible factors of –24 :
–1, 24 –2, 12 –3, 8 –4, 6
Since 3 – 8 = 5 , then a = 8 and b = –3
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Not-So-Special Trinomials

Trinomials with one negative term
– the constant
 Factor x2 + 5x – 24
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 are
changing the form of an expression ,
NOT solving an equation
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Not-So-Special Trinomials

Trinomials with one negative term
– the middle term
Sign of
both terms
Terms a , b of
same sign
 Solve x2 – 8x + 12 = 0 by factoring
 Find a and b such that
x2 + (a + b)x + ab = (x + a)(x + b)
Possible factors of 12 :
–1, –12 –2, –6 –3, –4
Since –2 – 6 = –8 , then a = –2 and b = –6
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Not-So-Special Trinomials

Trinomials with one negative term
– the middle term
 Solve x2 – 8x + 12 = 0 by factoring
Since –2 – 6 = –8 , then a = –2 and b = –6
This gives us:
x2 – 8x + 12 = (x – 2)(x – 6) = 0
So x – 2 = 0 or x – 6 = 0
Solution Set: { 2, 6 }
NOTE: This is a solution , since we are
solving an equation
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Not-So-Special Trinomials

Trinomials with no negative terms
Sign of
both terms
Terms a , b of
same sign
 Solve x2 + 11x + 28 = 0 by factoring
 Find a and b such that
x2 + (a + b)x + ab = (x + a)(x + b)
Possible factors of 28 :
1, 28 2, 14 4, 7
Since 4 + 7 = 11 , then a = 4 and b = 7
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Not-So-Special Trinomials

Trinomials with no negative terms
 Solve x2 + 11x + 28 = 0 by factoring
Since 4 + 7 = 11 , then a = 4 and b = 7
This gives us:
x2 + 11x + 28 = (x + 4)(x + 7) = 0
So x + 4 = 0 or x + 7 = 0
Solution Set: { – 4 , –7 }
NOTE: This is a solution , since we are
solving an equation
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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 ?
Original area:
A1 = 208 ft2
New area:
A = 2A2 = 216 ft2
9+x
Two methods:
A = (9 + x)(12 + x)
= 216
A = A1 + A2 + A3 + A4
= 216
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12 + x
12 ft
x ft
9 ft
A1 = 108
A2 = 9x
x ft
A3 = 12x
A 4 = x2
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Expanding a Room
Two methods:
OR
Either way:
So
x=
x=
A = (9 + x)(12 + x) = 216
A = A1 + A2 + A3 + A4 = 216
x2 + 21x + 108 = 216
x2 + 21x – 108 = 0
– 21 ±  (21)2 – 4(1)( –108)
2(1)
– 21 ±  876
2
≈ 4.3 sq. ft.
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Think about it !
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