This document defines and explains how to solve quadratic equations. A quadratic equation is an equation that can be written in the standard form ax^2 + bx + c = 0, where a, b, and c are real numbers and a is not equal to 0. There are several methods for solving quadratic equations covered in the document: factoring, taking the square root, completing the square, and using the quadratic formula. The discriminant, b^2 - 4ac, determines the number and type of solutions.

1. Quadratic Equations

2. Definition of a Quadratic Equation
• A quadratic equation in x is an equation
that can be written in the standard form
• ax2 + bx + c = 0
• where a, b, and c are real numbers with a
not equal to 0. A quadratic equation in x is
also called a second-degree polynomial
equation in x.

3. The Zero-Product Principle
If the product of two algebraic expressions is
zero, then at least one of the factors is equal
to zero.
If AB = 0, then A = 0 or B = 0.

4. Solving a Quadratic Equation by
Factoring
1. If necessary, rewrite the equation in the form
ax2 + bx + c = 0, moving all terms to one side,
thereby obtaining zero on the other side.
2. Factor.
3. Apply the zero-product principle, setting each
factor equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.

5. Text Example
• Solve 2x2 + 7x = 4 by factoring and then using the
zero-product principle.
Step 1 Move all terms to one side and obtain
zero on the other side. Subtract 4 from both sides
and write the equation in standard form.
2x2 + 7x - 4 = 4 - 4
2x2 + 7x - 4 = 0
Step 2 Factor.
2x2 + 7x - 4 = 0
(2x - 1)(x + 4) = 0

6. Solution cont.
• Solve 2x2 + 7x = 4 by factoring and then
using the zero-product principle.
Steps 3 and 4 Set each factor equal to
zero and solve each resulting equation.
2 x - 1 = 0 or x + 4 = 0
2 x = 1 x = -4
x = 1/2
Steps 5 check your solution

7. Example
(2x + -3)(2x + 1) = 5
4x2 - 4x - 3 = 5
4x2 - 4x - 8 = 0
4(x2-x-2)=0
4(x - 2)*(x + 1) = 0
x - 2 = 0, and x + 1 = 0
So x = 2, or -1

8. The Square Root Method
If u is an algebraic expression and d is a
positive real number, then u2 = d has exactly
two solutions.
If u2 = d, then u = Öd or u = -Öd
Equivalently,
If u2 = d then u = ±Öd

9. Completing the Square
If x2 + bx is a binomial, then by adding (b/2) 2,
which is the square of half the coefficient of
x, a perfect square trinomial will result.
That is,
x2 + bx + (b/2)2 = (x + b/2)2

10. Text Example
What term should be added to the binomial x2
+ 8x so that it becomes a perfect square
trinomial? Then write and factor the
trinomial.
The term that should be added is the square of
half the coefficient of x. The coefficient of x
is 8. Thus, (8/2)2 = 42. A perfect square
trinomial is the result.
x2 + 8x + 42 = x2 + 8x + 16 = (x + 4)2

11. Quadratic Equation
ax2 + bx + c = 0

12. Quadratic Formula
= - ± 2 - 4
x b b ac
a
2

13. x x
- + =
- - ± - -
=
( 8) ( 8) 4(1)(5)
= ± -
8 64 20
8 44
2
2
2(1)
8 5 0
2
2
= ±
x
x
x

14. = ±
8 2 11
2
= ±
2(4 11)
2
= ±
4 11
x
x
x

15. The Discriminant and the Kinds of Solutions
No x-intercepts
No real solution;
two complex imaginary
solutions
b2 – 4ac < 0
One x-intercept
One real solution
(a repeated solution)
b2 – 4ac = 0
Two x-intercepts
b2 – 4ac > 0 Two unequal real solutions
Graph of
y = ax2 + bx + c
Kinds of solutions
to ax2 + bx + c = 0
Discriminant
b2 – 4ac
to ax2 + bx +c = 0

16. The Pythagorean Theorem
The sum of the squares of the lengths of the
legs of a right triangle equals the square of
the length of the hypotenuse.
If the legs have lengths a and b, and the
hypotenuse has length c, then
a2 + b2 = c2

17. Quadratic Equations