1. Write 15x2 + 6x = 14x2 – 12 in standard form.
ANSWER
x2 + 6x +12 = 0
2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5.
ANSWER
–24
Classwork Answers…
4.8 (13-27 odd)
4.6 (multiples of 3)
 b  b 2  4ac
x
2a
When solving quadratic equations,
we’re looking for the x values where
the graph crosses the x axis
One of the methods we use to solve quadratic
equations is called the Quadratic Formula
 b  b  4ac
x
2a
2
Using the a, b, and c from ax2 + bx + c = 0
Must be equal to zero
EXAMPLE 1
Solve an equation with two real solutions
Solve x2 + 3x = 2.
x2 + 3x = 2
x2 + 3x – 2 = 0
x = – b + b2 – 4ac
2a
x = – 3 + 32 – 4(1)(–2)
2(1)
x = – 3 + 17
2
 3  17
x
 0.56
2
Write original equation.
Write in standard form.
Quadratic formula
a = 1, b = 3, c = –2
Simplify.
or
 3  17
x
 3.56
2
EXAMPLE 2
Solve an equation with one real solutions
Solve 25x2 – 18x = 12x – 9.
25x2 – 18x = 12x – 9.
Write original equation.
Write in standard form.
25x2 – 30x + 9 = 0.
x = 30 + (–30)2– 4(25)(9) a = 25, b = –30, c = 9
2(25)
30 + 0
x = 50
x = 53
ANSWER
The solution is 3
5
Simplify.
Simplify.
EXAMPLE 3
Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.
–x2 + 4x = 5
–x2 + 4x – 5 = 0.
x = –4 + 42 – 4(–1)(–5)
2(–1)
–4 + –4
x = –2
–4 + 2i
x=
–2
x=2+i
ANSWER
The solution is 2 + i and 2 – i.
Write original equation.
Write in standard form.
a = –1, b = 4, c = –5
Simplify.
Rewrite using the
imaginary unit i.
Simplify.
GUIDED PRACTICE
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
1.
x2
= 6x – 4
x2 – 6x + 4 = 0
a = 1 b = -6 c = 4
x = – b + b2 – 4ac
2a
6  (6) 2  4(1)(4)
x
2(1)
6  20
2
62 5
x
2
2(3  5 )
x
2
x
x  3 5
GUIDED PRACTICE
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
2.
4x2 – 10x = 2x – 9
4x2 – 12x + 9 = 0
a = 4 b = -12 c = 9
x = – b + b2 – 4ac
2a
12  (12) 2  4(4)(9)
x
2(4)
12 0
8
12
x
8
x
3
x
2
EXAMPLE 4
Use the discriminant
If the quadratic equation is in the standard form ax2 + bx + c = 0
The discriminant can be found using b2 – 4ac
If b2 – 4ac < 0
There are Two Imaginary solutions
If b2 – 4ac = 0
There is One Real solution
If b2 – 4ac > 0
There are Two Real solutions
Discriminant
Equation
b2 – 4ac
Solution(s)
a. x2 – 8x + 17 = 0
( –8)2 – 4(1)(17) = –4
b. x2 – 8x + 16 = 0
(–8)2 – 4(1)(16) = 0
One real
b. x2 – 8x + 15 = 0
(–8)2 – 4(1)(15) = 4
Two real
Two imaginary
GUIDED PRACTICE
for Example 4
Find the discriminant of the quadratic equation and give
the number and type of solutions of the equation.
4. 2x2 + 4x – 4 = 0
b2 – 4ac
5.
b2 – 4ac
(4)2 – 4(2)(-4) = 48
(12)2 – 4(3)(12) = 0
Two Real solutions
8x2 = 9x – 11
8x2 – 9x + 11 = 0
6.
b2 – 4ac
(-9)2
– 4(8)(11) = -271
Two Imaginary solutions
3x2 + 12x + 12 = 0
One Real solution
7.
4x2 + 3x + 12 = 3 – 3x
4x2 + 6x + 9 = 0
b2 – 4ac
(6)2 – 4(4)(9) = -108
Two Imaginary solutions
Classwork Assignment:
WS 4.8 (1-27 odd)