Using the Quadratic Equation Formula

(updated 18 January 2022)

One way to find the solutions to a quadratic equation is to use the quadratic formula:

x = [−b ± √(b² − 4ac)]/2a

The quadratic formula is used when factoring the quadratic expression (ax² + bx + c) is not easy or possible.

One requirement for using the formula is that a is not equal to zero (a ≠ 0), because the result would then be infinite (∞). Another requirement is that (b² > 4ac) to avoid imaginary solutions.

Besides having solutions consisting of rational numbers, solutions of quadratic equations can be irrational or even imaginary.

This lesson will answer those questions.

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Solving quadratic equations

You can find the values of x that solve the quadratic equation ax² + bx + c = 0 by using the quadratic formula, provided a, b, and c are whole numbers and a ≠ 0,

x = [−b ± √(b² − 4ac)]/2a

It is good to memorize the equation in words:

“x equals minus b plus-or-minus the square root of b-squared minus 4ac, divided by 2a.”

When not whole numbers

If a, b, or c are not whole numbers, you can multiply the equation by some value to make them whole numbers. For example, if the equation is:

x²/2 + 2x/3 + 1/6 = 0

Multiply both sides of the equal sign by 6, resulting in:

3x² + 4x + 1 = 0

This equation is then in the proper format for using the quadratic equation formula:

x = [−4 ± √(4² − 4*3*1)]/2*3

Rational solutions

Often the solutions to quadratic equations are rational numbers, which are integers or fractions.

The requirement for the solution to be an integer or fraction is that √(b² − 4ac) is a whole number.

Example 1

One example is the solution to the equation x² + 2x − 15 = 0. Substitute values in the formula:

x = [−b ± √(b² − 4ac)]/2a

a = 1, b = 2, and c = −15. Thus:

x = [−2 ± √(2² − 4*1*{−15})]/2

x = [−2 ± √(4 + 60)]/2

x = [−2 ± √(64)]/2

x = [−2 ± 8]/2

The two solutions are:

x = −10/2 and x = +6/2

x = −5 and x = 3

Example 2

Try the equation 2x² − x − 1 = 0:

x = [−b ± √(b² − 4ac)]/2a

x = [1 ± √(1² − 4*2*{−1})]/4

x = [1 ± √(1 + 8)]/4

x = [1 ± √(9)]/4

x = [1 ± 3)]/4

x = 4/4 and x = −2/4

Thus

x = 1 and x = −1/2

Irrational and Imaginary solutions

The solution to some quadratic equations consist of irrational values for x. In other words, the square root of b² − 4ac is not a whole number. For example, √2 is an irrational number equal to 1.41421… (where… means “and so on”).

An imaginary number is a multiple of √−1. It is called imaginary, since no number exists whose square is −1. Imaginary numbers are used in certain equations in electrical engineering, signal processing and quantum mechanics.

Irrational solution example

Consider the equation x² + 3x + 1 = 0:

x = [−b ± √(b² − 4ac)]/2a

x = [−3 ± √(3² − 4)]/2

x = [−3 ± √(9 − 4)]/2

x = [−3 ± √5]/2

x = −3/2 + (√5)/2 and x = −3/2 − (√5)/2

Both solutions are irrational numbers.

Imaginary solution example

Consider the equation x² + x + 1 = 0:

x = [−1 ± √(1² − 4)]/2

x = [−1 ± √−3]/2

x = −1/2 + (√−3)/2 and x = −1/2 − (√−3)/2

Both solutions are imaginary numbers.

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Use tools that make your work easier

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Resources and references

Websites

Quadratic Equation Calculator – From Math.com

Algebra Resources