Solving Quadratic Equations
1
Zero-Product Property
If the product of two factors is zero, then at least
one of the factors is 0
That is:
If ab = 0, then a = 0 or b = 0 or both a and b are 0
2
Example
Solve: (x – 5)(x + 4) = 0
Set each expression equal to 0
x–5=0
x=5
or
x+4=0
x=−4
The solution set is {{– 4, 5}
Check:
(x – 5)(x + 4) = 0
(x – 5)(x + 4) = 0
(5 – 5)(5 + 4) = 0
0(9) = 0
0=09
(– 4 – 5)(– 4 + 4) = 0
– 9(0) = 0
0=09
3
1
Factoring
a2 - b2
Special
Factors?
2 Terms
a3+b3
a3-b3
Perfect
Pe
fect
Trinomial
Factor Out
GCF
a=1 Trial &
Error
3 Terms
a≠1
a2+2ab+b2
a2-2ab+b2
Trial &
Error
Grouping
4 Terms
Grouping
Example
4
Find the break-even point
P ( x ) = 70 x − 10000 − 0.1x 2
Write in standard form P ( x ) = −0.1x 2 + 70 x − 10000
Factor out GCF
P ( x ) = −0.1( x 2 − 700 x + 100000 )
Factor trinomial
P ( x ) = −0.1( x − 200 )( x − 500 )
Break – even
means P(x) = 0
0 = −0.1( x − 200 )( x − 500 )
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Example
0 = −0.1( x − 200 )( x − 500 )
Solve for x
Find the break-even point
P ( x ) = 70 x − 10000 − 0.1x 2
x − 200 = 0 or x − 500 = 0
x = 200 or x = 500
We break even when we produce 200 or 500
units
6
2
Example
Find the break-even point
P ( x ) = 10000 − 70 x + 0.1x 2
Our x-intercepts are at 200 and 500
Practice
7
Solve each equation
using the Zero-Product
Property.
2x ( x − 4) = 0
2 x = 0 or x − 4 = 0
( x − 3)( x + 2 ) = 0
x − 3 = 0 or x + 2 = 0
x = 0 or x = 4
x = 3 or x = −2
{0,4}
{− 2,3}
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Square Root Method
The solutions to the quadratic equation x 2 = C are x = ± C
For example x 2 = 4, has the solutions x = −2 and x = 2
because (2)2 = 4 and (−2) 2 = 4
We take the square root of C. The primary root is C .
Since we use both C and − C we write ± C .
The solutions to the quadratic equation x 2 = C are x = ± C
9
3
Example
4 x 2 − 25 = 0
Write variable term on left side
and constant term on right side
4 x 2 = 25
Take the square root of each
side
2 x = ±5
Solve for x
x=±
Write solutions
⎧ 5 5⎫
⎨− , ⎬
⎩ 2 2⎭
5
2
10
Quadratic Formula
ax 2 + bx + c = 0
The quadratic formula x =
−b ± b 2 − 4ac
2a
Other useful information within the
quadratic formula
The vertex is at x =
−b
2a
The discrimant is b 2 − 4ac
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Example
3x 2 − 4 x − 2 = 0
Identify a, b, c
Use quadratic formula
Simplify x =
a = 3, b = −4, c = −2
x=
− ( −4 ) ±
( −4 ) − 4 ( 3)( −2 )
2 ( 3)
2
4 ± 16 + 24
4 ± 40
4 ± 2 10
=
=
6
6
6
4 2 10 2
10
= ±
= ±
6
6
3
3
Write solution
10 2
10 ⎪⎫
⎪⎧ 2
, +
⎨ −
⎬
3
3
3
3 ⎪⎭
⎩⎪
12
4
Complex Solutions
The solutions to x 2 = −C are x = ±i C
i 2 = −1 and ( −i ) = −1 and i = −1
2
13
Example
x2 − 5x + 7 = 0
a = 1, b = −5, c = 7
Identify a, b, c
Use quadratic formula
Simplify
Solution
x=
x=
− ( −5 ) ±
( −5 )
2 (1)
2
5 ± 25 − 28 = 5 ± −3
2
2
=
− 4 (1)( 7 )
5±i 3
2
⎧⎪ 5 − i 3 5 + i 3 ⎫⎪
,
⎨
⎬
2 ⎭⎪
⎩⎪ 2
14
Example
x2 − 5x + 7 = 0
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5
Completing the Square
1. Be sure that the coefficient of the highest
power is one.
If it is not, divide each term by that value to
g coefficient of one.
create a leading
2x2 − 8x + 6 = 0
x2 − 4 x + 3 = 0
16
Completing the Square
2. Move the constant term to the right hand side.
x2 − 4 x + 3 = 0
x 2 − 4 x = −3
17
Completing the Square
3. Prepare to add the needed value to create the
perfect square trinomial.
Be sure to balance the equation.
The boxes may help you remember to balance.
x 2 − 4 x = −3
x2 − 4 x +
= −3 +
18
6
Completing the Square
4. To find the needed value for the perfect square
trinomial, take half of the coefficient of the middle
term (x-term), square it, and add that value to
both sides of the equation.
x2 − 4 x +
= −3 +
⎛ −4 ⎞
⎜ ⎟ =4
⎝ 2 ⎠
2
x 2 − 4 x + 4 = −3 + 4
19
Completing the Square
5. Factor the perfect square trinomial.
x 2 − 4 x + 4 = −3 + 4
( x − 2)
2
=1
20
Completing the Square
6. Take the square root of each side and
solve.
Remember to consider both plus and minus
results.
results
( x − 2)
2
=1
x − 2 = ± 1 = ±1
x = 2 ±1
{1,3}
21
7
Example
Solve: x 2 – 6x – 7 = 0
Add 7 to both sides
x 2 – 6x = 7
Complete the square
x 2 – 6x + 9 = 7 + 9
Factor
(x – 3) 2 = 16
x − 3 = ± 16
Use the Square Root Property
x − 3 = ±4
Simplify
x = 3± 4
Add 3 to both sides
{−1,7}
Solution
22
The Discriminant
In the quadratic formula
x=
−b ± b 2 − 4ac
2a
The quantity b 2 – 4ac is called the discriminant of the
quadratic equation, because its value tells us the
number of solutions and the type of solution
The Discriminant
Discriminant
Number of
Solutions
Type of
Solutions
Positive
2
Real
Zero
1
(Repeated Root)
Real
Negative
2
Complex
(Not real)
8
2 Real Solutions
Recommend: Solve(, factoring, graphing, quadratic
formula, or completing the square
25
1 Real Solution
Recommend: Solve(, factoring, graphing, quadratic
formula, or square root method
26
No Real Solutions
Recommend: cSolve(, quadratic formula, or
completing the square
27
9
Tutorials Online
These concepts are important and we must master
them. Easily solving quadratic equations requires
practice.
Try the Study Guide at MyMathLab. Some students
find the following websites useful:
Purple Math
http://www.purplemath.com/modules/index.htm
Virtual Math Lab at TAMU
http://www.wtamu.edu/academic/anns/mps/math/
mathlab/col_algebra/index.htm
Library of Math
http://www.libraryofmath.com/
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