Lesson 5: The Quadratic Formula

New Concept

The quadratic formula is your master key for solving any quadratic equation, $ax^2 + bx + c = 0$. It finds the exact solutions—even when factoring doesn't work—and tells us the precise x-intercepts of the equation's graph.

What’s next

Now that you have the formula, you'll apply it in interactive examples and practice problems to find exact solutions and x-intercepts of parabolas.

Property

The solutions of the equation

are given by the formula

Property

Before we use the quadratic formula, we must write the equation in standard form, $ax^2 + bx + c = 0$, so that we can identify the coefficients $a$, $b$, and $c$. If the coefficients are fractions, it helps to clear the fractions by multiplying each term by the LCD.

Examples

• To solve $5x^2 = 8x - 3$, first rearrange it to standard form: $5x^2 - 8x + 3 = 0$. Now you can correctly identify $a=5, b=-8,$ and $c=3$.

• Given the equation $x^2 - \frac{x}{3} = \frac{4}{3}$, first multiply by the LCD, 3, to get $3x^2 - x = 4$. Then, rewrite in standard form as $3x^2 - x - 4 = 0$, so $a=3, b=-1, c=-4$.

Property

If the value under the radical in the quadratic formula, $b^2 - 4ac$, is a negative number, the equation has no real-valued solutions. This is because the square root of a negative number is not a real number. The solutions are a type of number called complex numbers.

Examples

• For $x^2 - 4x + 5 = 0$, the formula gives $x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2} = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm \sqrt{-4}}{2}$. Since $\sqrt{-4}$ is not a real number, there are no real solutions.

• To solve $x^2 + 25 = 0$, the formula with $a=1, b=0, c=25$ gives $x = \frac{0 \pm \sqrt{0^2 - 4(1)(25)}}{2} = \frac{\pm \sqrt{-100}}{2}$. There are no real solutions.

Property

The $x$-intercepts of the graph of $y = ax^2 + bx + c$ are the solutions of $ax^2 + bx + c = 0$. There are three possibilities:

1. If both solutions are real numbers, and unequal, the graph has two $x$-intercepts.

2. If the solutions are real and equal, the graph has one $x$-intercept, which is also its vertex.