1. Quadratic Equations and Functions
Chapter 8
Quadratic Equations and Functions

2. Solving Quadratic Equations by Completing the Square
§ 8.1
Solving Quadratic Equations by Completing the Square

3. Square Root Property
We previously have used factoring to solve quadratic equations.
This chapter will introduce additional methods for solving quadratic equations.
Square Root Property
If b is a real number and a2 = b, then

4. Square Root Property Example: Solve x2 = 49 Solve 2x2 = 4 x2 = 2
Solve (y – 3)2 = 4
y = 3  2
y = 1 or 5

5. Square Root Property Example: Solve x2 + 4 = 0 x2 = 4
There is no real solution because the square root of 4 is not a real number.

6. Square Root Property Example: Solve (x + 2)2 = 25 x = 2 ± 5
x = 2 + 5 or x = 2 – 5
x = 3 or x = 7

7. Square Root Property
Example:
Solve (3x – 17)2 = 28
3x – 17 =

8. Completing the Square
In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left.
Also, the constant on the left is the square of the constant on the right.
So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).

9. Completing the Square Example: x2 – 10x x2 + 16x x2 – 7x
What constant term should be added to the following expressions to create a perfect square trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
add

10. Completing the Square Example:
We now look at a method for solving quadratics that involves a technique called completing the square.
It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.

11. Completing the Square
Solving a Quadratic Equation in x by Completing the Square
If the coefficient of x2 is 1, go to Step 2. Otherwise, divide both sides of the equation by the coefficient of x2.
Isolate all variable terms on one side of the equation.
Complete the square for the resulting binomial by adding the square of half of the coefficient of the x to both sides of the equation.
Factor the resulting perfect square trinomial and write it as the square of a binomial.
Use the square root property to solve for x.

12. Solving Equations Example: Solve by completing the square.
y2 + 6y = 8
y2 + 6y + 9 = 8 + 9
(y + 3)2 = 1
y + 3 = ± = ± 1
y = 3 ± 1
y = 4 or 2

13. Solving Equations Example: (y + ½)2 = Solve by completing the square.
y2 + y – 7 = 0
y2 + y = 7
y2 + y + ¼ = 7 + ¼
(y + ½)2 =

14. Solving Equations Example: Solve by completing the square.
2x2 + 14x – 1 = 0
2x2 + 14x = 1
x2 + 7x = ½
x2 + 7x = ½ =
(x + )2 =

15. Solving Quadratic Equations by Using the Quadratic Formula
§ 8.2
Solving Quadratic Equations by Using the Quadratic Formula

16. The Quadratic Formula
Another technique for solving quadratic equations is to use the quadratic formula.
The formula is derived from completing the square of a general quadratic equation.

17. The Quadratic Formula Quadratic Formula
A quadratic equation written in the form, ax2 + bx + c = 0, has the solutions

18. The Quadratic Formula Example:
Solve 11n2 – 9n = 1 by the quadratic formula.
11n2 – 9n – 1 = 0, so
a = 11, b = -9, c = -1

19. The Quadratic Formula Example:
Solve x2 + x – = 0 by the quadratic formula.
x2 + 8x – 20 = (multiply both sides by 8)
a = 1, b = 8, c = -20

20. The Quadratic Formula Example:
Solve x(x + 6) = 30 by the quadratic formula.
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
So there is no real solution.

21. The Discriminant
The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is positive, 0, or negative.
The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

22. The Discriminant Example:
Use the discriminant to determine the number and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions.