1. Warm Up #4 1. Write 15x2 + 6x = 14x2 – 12 in standard form. ANSWER
2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5.
ANSWER
–24

3. 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
Using the a, b, and c from ax2 + bx + c = 0
Must be equal to zero

4. Solve an equation with two real solutions
EXAMPLE 1
Solve an equation with two real solutions
Solve x2 + 3x = 2.
x2 + 3x = 2
Write original equation.
x2 + 3x – 2 = 0
Write in standard form.
x =
– b + b2 – 4ac 2a
Quadratic formula
x =
– – 4(1)(–2)
2(1)
a = 1, b = 3, c = –2
x =
– 3 + 17 2
Simplify. or

5. Solve an equation with one real solutions
EXAMPLE 2
Solve an equation with one real solutions
Solve 25x2 – 18x = 12x – 9.
25x2 – 18x = 12x – 9.
Write original equation.
25x2 – 30x + 9 = 0.
Write in standard form.
x =
(–30)2– 4(25)(9)
2(25)
a = 25, b = –30, c = 9
x = 50
Simplify.
x = 3 5
Simplify. 3 5
The solution is
ANSWER

6. Solve an equation with imaginary solutions
EXAMPLE 3
Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.
–x2 + 4x = 5
Write original equation.
–x2 + 4x – 5 = 0.
Write in standard form.
x =
– 42 – 4(–1)(–5)
2(–1)
a = –1, b = 4, c = –5
x =
– –4 –2
Simplify.
–4 + 2i
x = –2
Rewrite using the imaginary unit i.
x = 2 + i
Simplify.
The solution is 2 + i and 2 – i.
ANSWER

7. GUIDED PRACTICE
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
x =
– b + b2 – 4ac 2a
x2 = 6x – 4
x2 – 6x + 4 = 0
a = 1 b = -6 c = 4

8. GUIDED PRACTICE
for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
4x2 – 10x = 2x – 9
x =
– b + b2 – 4ac 2a
4x2 – 12x + 9 = 0
a = 4 b = -12 c = 9

9. 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 < There are Two Imaginary solutions
If b2 – 4ac = There is One Real solution
If b2 – 4ac > There are Two Real solutions
Discriminant
Equation
b2 – 4ac
Solution(s)
a. x2 – 8x + 17 = 0
( –8)2 – 4(1)(17) = –4
Two imaginary
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

10. GUIDED PRACTICE
for Example 4
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
3x2 + 12x + 12 = 0
2x2 + 4x – 4 = 0
b2 – 4ac
b2 – 4ac
(4)2 – 4(2)(-4) = 48
(12)2 – 4(3)(12) = 0
Two Real solutions
One Real solution
8x2 = 9x – 11 7.
4x2 + 3x + 12 = 3 – 3x
8x2 – 9x + 11 = 0
4x2 + 6x + 9 = 0
b2 – 4ac
b2 – 4ac
(6)2 – 4(4)(9) = -108
(-9)2 – 4(8)(11) = -271
Two Imaginary solutions
Two Imaginary solutions