```QUADRATIC
EQUATIONS
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Theresa Hert
Radicals with index 2 are referred
to as square roots.
Break down the radicand, the number inside
Circle a pair of matching factors,
take out THE factor.
Since no operation sign is visible, the “glue”
holding everything together is Multiplication.
When you bring a factor out of the radical, it gets
multiplied to the number in front of the radical.
5 63
5 337
53 7
15 7
Simplify Rational Expressions
To reduce the fraction, Factor.
Plus sign – use one set of parentheses
to factor out what is common.
Simplify this Rational Expression
9  45
6
9 3 3 5  93 5
6
6

3 3 1 5
6

3 5

2
contain both an equal sign and
a variable with exponent 2.
General form: ax2 + bx + c = 0
• A quadratic equation is an equation
equivalent to an equation of the type
ax2 + bx + c = 0, where a is nonzero
• We can solve a quadratic equation by
 b  b  4ac
x
2a
2
• Solve the equation ax2 + bx + c = 0 for x
by Completing the Square
ax  bx   c
b
c
x  x
a
a
b
b
c
b
x  x
 
a
4a
a 4a
2
2
2
2
2
2
2
b
b
4ac
b
x  x


a
4a
4a
4a
b  b  4ac

x  
2a 
4a

b 
b  4ac

x  
2a 
4a

b
b  4ac
x 
2a
4a
2
2
2
2
2
2
2
2
2
2
2
2
2
2
• The solution to the equation ax2 + bx + c = 0 is
 b  b  4ac
x
2a
2
6y2 – 3y – 5 = 0
a = 6 b = -3 c = -5
(3)  (3)  4(6)(5)
x
2(6)
2
3  9  120
x
12
6y2 – 3y – 5 = 0
a = 6 b = -3 c = -5
(3)  (3)  4(6)(5)
x
2(6)
3  129
3  9  120
x
x
12
12
2
you can NOT reduce the fraction
Ex: Use the Quadratic Formula to solve1x2 + 7x
7 +6=0
Recall: For quadratic equation ax2 + bx + c = 0,
the solutions to a quadratic equation are given by
 b  b 2  4ac
x
2a
Identify a, b, and c in ax2 + bx + c = 0:
a=1
b= 7
c= 6
Now evaluate the quadratic formula at the identified
values of a, b, and c
 7  7  4(1)(6)
x
2(1)
2
x2 + 7x + 6 = 0
a=1
b=7 c=6
 7  49  24
x
2
 7  25
x
2
75
x
2
x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6
x = { - 1, - 6 }
Ex: Use the Quadratic Formula to solve
2
2m2 + 1m – 10 = 0
Recall: For quadratic equation ax2 + bx + c = 0,
the solutions to a quadratic equation are given by
 b  b 2  4ac
m
2a
Identify a, b, and c in am2 + bm + c = 0:
a=2
b= 1
c = - 10
Now evaluate the quadratic formula at the identified
values of a, b, and c
 1  1  4(2)(10)
m
2(2)
2
2x2 + 1x – 10 = 0
a=2
b = 1 c = -10
 1  1  80
m
4
 1  81
m
4
1 9
m
4
m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2
m = { 2, - 5/2 }
Ex: Use the Quadratic Formula to solve
x2 + 5x = -3
1 x2 + 55x + 3 = 0
Identify a, b, and c in ax2 + bx + c = 0:
a=1
b=+5
c= 3
Now evaluate the quadratic formula at the identified
values of a, b, and c
 b  b  4ac
x
2a
2
5  5  4( 1 )( 3 )
x
2( 1 )
2
x2 + 5x + 3 = 0
a=1 b=5 c=3
5  25  12
x
2
5  13
x
2
5  13
x
2
and
5  13
x
2
Ex: Use the Quadratic Formula to solve
10x2 – 5x = 0
10
10x2 –- 5x
5 + 00 = 0
Identify a, b, and c in ax2 + bx + c = 0:
a = 10
b=-5
c= 0
Now evaluate the quadratic formula at the identified
values of a, b, and c
 b  b  4ac
x
2a
2
x
  5  
 5   4  10  0 
2  10 
2
5  25  0
x
20
55
x
20
1
x
and
2
10x2 – 5x + 0 = 0
x0
a = 10 b = -5 c = 0
2x 2  6 x  1  0
a2
  6  
b  6
 6   4  2 1
2 2
6  36  8
4
6  28
4
c 1
2
6  2  2 7
4
62 7
4

2 3 7
4
3 7
2

```