# Section 1.4 - Quadratic Equations

Section Objectives

1. Solve quadratic equations by factoring.
2. Solve quadratic equations by using square roots.

A quadratic equation in the variable $x$ is an equation of the form $ax^2+bx+c=0$, where $a \ne 0$. This is an example of a polynomial equation. In fact, it is a 2nd degree polynomial equation.

There are three common approaches to solving quadratic equations:

1. Factor and set each factor to zero.
2. Use square roots to "undo" the square.

### Using Factoring to Solve Quadratic Equations

If several numbers (or expressions) are multiplied and their product is zero, then one of the factors must be zero. We normally state this property by saying:

If $AB=0$, then $A=0$ or $B=0$.

This property allows us to solve quadratic (or higher-order) equations by factoring.

#### Examples

• Solve for $x$: $\quad (2x+1)(x-5)=0$

• Solve for $x$: $\quad 3x(x-1)(x+3)(x-9)=0$

• Solve for $x$ by factoring: $\quad x^2-2x-3=0$

• Solve for $v$ by factoring: $\quad v^2+v-2=0$

• Solve for $u$ by factoring: $\quad 5u^2-6=-13u$

• Solve for $w$ by factoring: $\quad 4w^2+7w=-3$

### Using Square Roots to Solve Quadratic Equations

A quadratic equation that has the form $X^2=a$ can be solved by taking square roots of both sides of the equation. Remember that a perfect square usually has two square roots, one positive and one negative.

If $X^2=a$, then $X=\sqrt{a}$ or $X = -\sqrt{a}$.

#### Examples

• Solve for $x$ by using square roots: $\quad x^2+3=12$

• Solve for $w$ by using square roots: $\quad (w-6)^2=49$

• Solve for $x$ by using square roots: $\quad (2x+3)^2 = 0$

The method of square roots can be used to derive a formula that is capable giving the solutions to any quadratic equation. This formula is called the quadratic formula:

If $ax^2+bx+c=0$, then $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

The quantity under the radical, $b^2-4ac$, is called the discriminant. Its sign tells us about the solutions:

1. If $b^2-4ac=0$, then there is only one real number solution.
2. If $b^2-4ac>0$, then there are two different real number solutions.
3. If $b^2-4ac<0$, then there are two different complex number solutions.