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Using the Quadratic Equation Formula

by Ron Kurtus (revised 16 December 2015)

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.

Questions you may have include:

This lesson will answer those questions.



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.

Summary

The quadratic formula is used when the solution to a quadratic equation cannot be readily solved by factoring. It is worthwhile to memorize the quadratic formula. Besides having solutions consisting of rational numbers, solutions of quadratic equations can be irrational or even imaginary.


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