Prepared by Doron Shahar
Quadratic Equations
Prepared by Doron Shahar
Warm-up: page 15
A quadratic equation is an equation that can be written in the
2
ax
 bx  c  0 where a, b, and c are constants and a ≠ 0.
form _____________
The zero product property says that if AB  0 ,
B0
A  0 or ________.
then either ________
FOIL: ( x  2)(x  3)
Factor: x  9 x  14
2
Solve for x : x  4
2
Prepared by Doron Shahar
FOIL and Factoring
FOIL
Factor
First
2
Outside ( x  2)(x  3)
x

9
x

14
Inside
Last
7  2  9 and 7  2  14
2
x  3x  2 x  6
x  x6
2
x  (2  3) x  (2)(3)
2
( x  7)(x  2)
Prepared by Doron Shahar
1.4.1 Solve by Factoring
Starting Equation
Factor
Zero Product Property
Solution
x  9 x  14  0
2
( x  7)(x  2)  0
( x  7)  0 or ( x  2)  0
x  7 or x  2
Prepared by Doron Shahar
1.4.2 Solve by Factoring
Starting Equation
Factor
Zero Product Property
Solution
x  8x  9  0
2
( x  1)(x  9)  0
( x  1)  0 or ( x  9)  0
x  1 or x  9
Prepared by Doron Shahar
Solve by Factoring
Starting Equation
( x  2)(x  3)  6
x  x6  6
2
FOIL
6
6
x  x  12  0
( x  4)(x  3)  0
2
Factor
Zero Product Property
Solution
( x  4)  0 or ( x  3)  0
x  4 or x  3
Prepared by Doron Shahar
Intro to Completing the square
Starting Equation
Take square root of both
sides of the equation
x 4
2
x 
2
4
Place ± on the right side
of the equation
x 4
Solution
x  2
Prepared by Doron Shahar
Intro to Completing the square
2
( x  1)  6
Starting Equation
Take square root
( x  1)  6
2
Insert ± on right side
x 1   6
Group like terms
x  1 6
Solution
1
1
x  1  6 or x  1  6
Prepared by Doron Shahar
Goal of Completing the square
The goal of completing the squares is to get a quadratic
equation into the following form:
Starting Equation
Take square root
( x  h)  k
2
Solution
( x 1)  6
( x  h)  k
Insert ± on right side
Group like terms
eg.
2
2
xh   k
h
h
x  h  k
x  h  k or x  h  k
Prepared by Doron Shahar
1.4.5 Completing the square
The goal of completing the squares is to get a quadratic
2
equation into the following form: ( x  h)  k
Starting Equation
Add 3 to both sides
Multiply both sides by 2
Desired form
1
2
( x 1) 3  0
2
3
3
1
2
( x  1)  3
2
2  ( x 1)  2  3
1
2
2
( x  1)  6
2
Prepared by Doron Shahar
Example: Completing the square
Starting Equation
Add 8 to both sides
Multiply both sides by 2
Desired form
1
2
( x  3) 8  0
1
2
2
8
8
( x  3)  8
2
2  ( x  3)  2  8
1
2
2
( x  3)  16
2
Prepared by Doron Shahar
Completing the square
Equation from
previous slide
Take square root
Insert ± on right side
Group like terms
Solution
( x  3)  16
2
( x  3)  16
2
x  3  4
3
3
x  3 4
x  7 or x  1
Prepared by Doron Shahar
1.4.2 General method of
Completing the square
Starting Equation
x  8x  9  0
Add 9 to both sides
x  8x  9
2
2
Add (−8/2)2=16
to both sides
x  8x  16  9  16
Factor left side
( x  4)  25
2
2
Prepared by Doron Shahar
1.4.4 General method of
Completing the square
3x  2 x  8
2
Starting Equation
x  x
2
Divide both sides by 3
Add ((−2/3)/2)2=1/9
to both sides
Factor left side
2
3
8
3
x  x  
2
2
3
1
9
(x  ) 
1 2
3
8
3
25
9
1
9
Prepared by Doron Shahar
1.4.3 General method of
Completing the square
x  6 x  22  0
2
Starting Equation
Subtract 22 from both sides
Add (−6/2)2=9
to both sides
Factor left side
x  6 x  22
2
x  6 x  9  22  9
2
( x  3)  13
2
Prepared by Doron Shahar
No solutions in quadratic equation
2
Try solving
( x  3)  13
Take square root
( x  3)   13
2
¡PROBLEMA! You CANNOT take the square root of a
negative number. Therefore, the equation has no solution.
Solving by factoring works only if the equation has a solution.
Completing the square always works, and can be used to
determine whether a quadratic equation has a solution.
Prepared by Doron Shahar
General method of
Completing the square
ax  bx  c  0
2
Starting Equation
ax  bx  c
2
Subtract c from both sides
x  x
2
Divide both sides by a
Add ((b/a)/2)2=(b/2a)2
to both sides
Factor left side
x  x
2
(x 
b
a
b
a
     
b 2
2a
c
a
)  
b 2
2a
c
a
c
a
b 2
2a
 
b 2
2a
Prepared by Doron Shahar
Quadratic Formula
If we solve for x in the previous equation, ( x 
)  
we get an equation called the quadratic formula.
b 2
2a
c
a
 ,
b 2
2a
 b  b  4ac
x
2a
2
Quadratic Formula
The quadratic formula gives us the solutions to every quadratic equation.
2
Starting Equation
ax  bx  c  0
 b  b  4ac
x
2a
2
Solution
Prepared by Doron Shahar
Quadratic Formula Song
 b  b  4ac
x
2a
2
Quadratic Formula
Please sing along.
Prepared by Doron Shahar
Using the quadratic formula
9x  6x 1  0
2
Starting Equation
a9
b6
c 1
ax  bx  c  0
2
Plug 9 in for a, 6 for b, and 1 for c in the quadratic formula.
Quadratic Formula
Solution
x
b
b  4ac
2a
2
 (6)  (6) 2  4(9)(1)
x
2(9)
Prepared by Doron Shahar
Simplify your solution
Simplify
 (6)  (6) 2  4(9)(1)
x
2(9)
6 0
x
18
60
x
18
6
x
18
Solution
1
x
3
Prepared by Doron Shahar
1.4.1 Using the quadratic formula
Starting Equation
a 1
b9
c  14
1 x  9 x  14  0
2
ax  bx  c  0
2
Plug 1 in for a, 9 for b, and 14 for c in the quadratic formula.
Quadratic Formula
Solution
x
b
b  4ac
2a
2
 (9)  (9) 2  4(1)(14)
x
2(1)
Prepared by Doron Shahar
Simplify your solution
Simplify
x
 (9) 
(9) 2  4(1)(14 )
2(1)
 9  25
x
2
95
x
2
95
95
x
or x 
2
2
Solution
x  2 or x  7
Prepared by Doron Shahar
1.4.3 Using the quadratic formula
2
+( 6)x  22  0
Starting Equation
1x
a  1 b  6 c  22
ax  bx  c  0
2
Plug 1 in for a, −6 for b, and 22 for c in the quadratic formula.
Quadratic Formula
Solution
x
b
b  4ac
2a
2
 (6)  (6) 2  4(1)(22)
x
2(1)
Prepared by Doron Shahar
Simplify your solution
Simplify
 (6)  (6) 2  4(1)(22)
x
2(1)
x
6
 52
2
You cannot take the square root of a negative number.
Therefore, there is no solution.
No Solution
Prepared by Doron Shahar
Discriminant
Equation
Discriminant # of Solutions
ax  bx  c  0 b 2  4ac
2
9x  6x 1  0
1 Solut ion if b 2 4ac  0
2 Solut ionsif b 2 4ac  0
No Solut ionsif b 2 4ac  0
2
x  9 x  14  0
2
x  6 x  22  0
2
(6)  4(9)(1)  0
2
1 Solution
(9)  4(1)(14)  25  0 2 Solutions
2
No
2
(6)  4(1)(22)  52  0 Solutions
Prepared by Doron Shahar
Calculator
Put the Quadratic Formula program on your calculator.
 Instructions are in the back of the class notes.
OR
 You can come in to office hours to have me load the
program onto your calculator.
Warning!
The calculator will not always give you exact answers.
Prepared by Doron Shahar
Simplifying expressions with
If x  0, then x  x
2
25 
ab  a b
5 5
2
18  9 2  9 2  3 2
20  4 5 
4 5 2 5
6  20 6  2 5 6 2 5

 
 3 5
2
2
2
2
Prepared by Doron Shahar
Quadratic equations with Decimals
 If a quadratic equation has decimals, it is easiest to simply use
the quadratic formula. If you want, you can multiply both sides
of the equation by a power of 10 (i.e., 10, 100, 1000, etc) to get
rid of the decimals. This can make it easier to simplify the answer
if you are evaluating the quadratic formula without a calculator.
Prepared by Doron Shahar
Quadratic equations with Fractions
 If a quadratic equation has fractions (and does not factor), it is
often easy to simply use the quadratic formula. If you want, you
can multiply both sides of the equation by the least common
denominator to get rid of the fractions. This can make it easier
to simplify the answer.
 If a quadratic equation has fractions (and factors), it is often
easier to factor after having gotten rid of the fractions.
Prepared by Doron Shahar
Variables in the denominators
 If an equation has variables in the denominator, it is NOT a
quadratic equation. Such equations, however, can lead to linear
equations.
 We treat such equations like those with fractions. That is, we
multiply both sides of the equation by a common denominator
to get rid of the variables in the dominators. Ideally, we should
multiply by the least common denominator.
Example:
 If our problem has B +16 in the denominator of one term, and
B in the denominator of another term, we multiply both sides
of the equation by B(B+16). After the multiplication, the terms
will have no variables in the denominators.
Prepared by Doron Shahar
Variables in the denominators
Starting Equation
Multiply both sides
by B(B+16)
Distribute the
B(B+16)
15
15
 1
B  16 B
15
 15
B( B  16) 
   B( B  16)1
 B  16 B 
 15 
15
B( B  16) 

B
(
B

16
)
 B( B  16)



 B  16
B
15B  15( B  16)  B( B  16)
The problem is now in a form you can solve.
Prepared by Doron Shahar
Systems of equations
There are two methods for solving systems of equations:
Substitution and Elimination. Both work by combining
the equations into a single equation with one variable.
And sometimes the resulting equation leads to a
quadratic equation.
We will only review substitution, because elimination is
not a common method when working with systems of
equations that lead to quadratic equations.
Prepared by Doron Shahar
Substitution
Starting Equations
Substitute B+16 for J
in the second equation
Solution for B
J  B  16
15 15

1
J
B
15
15

1
B  16
B
B  24 or B  10
Plug in 24 and −10 for
B in the first equation
to get the solution for J
J  24  16  40 or
J  10  16  6