1. Algebra 1
Section 12.6

2. Solving Quadratic Equations
Choose the method that is most appropriate for the given problem.
Square root property
Factoring
Quadratic formula

3. Solving Quadratic Equations
Completing the square is usually not used since one of the other techniques is usually easier.
However, it is valuable in other applications you will use later.

4. Solving Quadratic Equations
Write the quadratic equation in the form ax2 + bx + c = 0.
If b = 0, isolate x2 and apply the Square Root Property.

5. Solving Quadratic Equations
If the polynomial factors easily, solve the equation by factoring.
Otherwise, use the quadratic formula:
x =
-b ± b2 – 4ac 2a

6. The Discriminant
The radicand in the quadratic formula, b2 – 4ac, determines the number and type of solutions to the equation.
This value, D = b2 – 4ac, is called the discriminant.

7. The Discriminant Number and type of solutions D = b2 – 4ac
D > 0 and a perfect square
2 rational
D > 0 but not a perfect square
2 irrational

8. Number and type of solutions
The Discriminant
Number and type of solutions
D = b2 – 4ac
D = 0
1 rational
no real solutions
D < 0

9. Example 1a x2 + 5x – 7 = 0 a = 1, b = 5, c = -7 D = b2 – 4ac
= 52 – 4(1)(-7) = 25 – (-28)
D = 53
There are 2 irrational solutions.

10. Example 1b x2 = 156 – x x2 + x – 156 = 0 a = 1, b = 1, c = -156
D = b2 – 4ac
= 12 – 4(1)(-156) = 625
There are 2 rational solutions.

11. Example 1c 3x2 + 5x = -9 3x2 + 5x + 9 = 0 a = 3, b = 5, c = 9
D = b2 – 4ac
= 52 – 4(3)(9) = -83
There are no real solutions.

12. Example 2a If the solutions are x = 5 and x = -7, then we know...
...and using the Multiplication Property of Equality, we get:
x2 + 2x – 35 = 0
(x – 5)(x + 7) = 0 • 0

13. Example 2b x = ⅔ and x = -¼ Multiply to clear the fractions:
(3x – 2)(4x + 1) = 0 • 0
12x2 – 5x – 2 = 0

14. Example 3 Let x = hypotenuse length x – 3 = length of one leg
x – 5 = length of other leg
Since a2 + b2 = c2,
(x – 5)2 + (x – 3)2 = x2

15. The solutions will be irrational.
Example 3
(x – 5)2 + (x – 3)2 = x2
x2 – 10x x2 – 6x + 9 = x2
x2 – 16x + 34 = 0
D = b2 – 4ac
D = (-16)2 – 4(1)(34) = 120
The solutions will be irrational.

16. Example 3
x2 – 16x + 34 = 0
D = 120 8 1
x =
16 ± 120 2 =
16 ± 2 1
x = 8 ± 30

17. Example 3
x = 8 ± 30
x =
≈ ft
x = 8 – 30
≈ 2.52 ft

18. Example 3 Let x = hypotenuse length x – 3 = length of one leg
x – 5 = length of other leg
Hypotenuse ≈ ft
Leg1 = x – 5 ≈ 8.48 ft
Leg2 = x – 3 ≈ ft

19. Homework: pp