1. 7.4 The Quadratic Formula and the Discriminant
Algebra 2
Mrs. Spitz
Spring 2007

2. Objectives Solve equations using the quadratic formula.
Use the discriminant to determine the nature of the roots of a quadratic equation.

3. Assignment
pp #7-33 odd

4. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
General form of a quadratic equation.
Divide all by a
Simplify
Subtract c/a on both sides.
Multiply by ½ and square the result.

5. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
Add the result to both sides.
Simplify
Multiply by common denominator
Simplify

6. Derive the quadratic formula from ax2 + bx + c = 0 a≠ 0
Square root both sides
Simplify
Common denominator/subtract from both sides
Simplify

7. Quadratic Formula
The solutions of a quadratic equation of the form ax2 + bx + c with a ≠ 0 are given by this formula:
MEMORIZE!!!!

8. Ex. 1: Solve t2 – 3t – 28 = 0 a = 1 b = -3 c = -28
There are 2 distinct roots—Real and rational.

9. Ex. 1: Solve t2 – 3t – 28 = 0 CHECK: t2 – 3t – 28 = 0
72 – 3(7) – 28 = 0
49 – 21 – 28 = 0
49 – 49 = 0 
CHECK:
t2 – 3t – 28 = 0
(-4)2 – 3(-4) – 28 = 0
– 28 = 0
28 – 28 = 0 

10. Ex. 1: Solve t2 – 3t – 28 = 0 -- GRAPH

11. Ex. 2: Solve x2 – 8x + 16 = 0 a = 1 b = -8 c = 16
There is 1 distinct root—Real and rational.

12. Ex. 2: Solve x2 – 8x + 16 = 0 CHECK: x2 – 8x + 16 = 0
(4)2 – 8(4) + 16 = 0
16 – = 0
32 – 32 = 0 
There is 1 distinct root—Real and rational.

13. Ex. 2: Solve Solve x2 – 8x + 16 = 0 -- GRAPH

14. Ex. 3: Solve 3p2 – 5p + 9 = 0 a = 3 b = -5 c = 9
There is 2 imaginary roots.

15. Ex. 3: Solve 3p2 – 5p + 9 = 0
NOTICE THAT THE PARABOLA DOES NOT TOUCH THE X-AXIS.

16. Note:
These three examples demonstrate a pattern that is useful in determining the nature of the root of a quadratic equation. In the quadratic formula, the expression under the radical sign, b2 – 4ac is called the discriminant. The discriminant tells the nature of the roots of a quadratic equation.

17. Value of the discriminant
The discriminant will tell you about the nature of the roots of a quadratic equation.
Equation
Value of the discriminant
Roots
Nature of roots
t2 – 3t – 28 = 0
b2 – 4ac =
(-3)2 – 4(1)(-28) = 121
{7, - 4}
2 real roots
x2 – 8x + 16 = 0
(-8)2 – 4(1)(16) = 0
{0}
1 real root
3p2 – 5p + 9 = 0
(-5)2 – 4(3)(9) = -83
2 imaginary roots

18. 2x2 + x – 3 = 0 a = 2 b = 1 c = -3 b2 – 4ac = (1)2 – 4(2)(-3) = 1 + 24
Ex. 4: Find the value of the discriminant of each equation and then describe the nature of its roots.
2x2 + x – 3 = 0
a = 2 b = 1 c = -3
b2 – 4ac = (1)2 – 4(2)(-3) =
= 25
The value of the discriminant is positive and a perfect square, so 2x2 + x – 3 = 0 has two real roots and they are rational.

19. x2 + 8 = 0 a = 1 b = 0 c = 8 b2 – 4ac = (0)2 – 4(1)(8) = 0 – 32 = – 32
Ex. 5: Find the value of the discriminant of each equation and then describe the nature of its roots.
x2 + 8 = 0
a = 1 b = 0 c = 8
b2 – 4ac = (0)2 – 4(1)(8)
= 0 – 32
= – 32
The value of the discriminant is negative, so x2 + 8 = 0 has two imaginary roots.