Solving Quadratic Equations Using the Quadratic Formula

Solving Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes.’ In this section, we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’ so that we would do the algebraic steps only once and then use the new formula to find the value of the specific variable. Now, we will go through the steps of completing the square in general to solve a quadratic equation for x. It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form \(a{x}^{2}+bx+c=0\) as you read through the algebraic steps below, so you see them with numbers as well as ‘in general.’

\(\begin{array}{cccccc}\text{We start with the standard form of a quadratic equation}\hfill & & & \hfill a{x}^{2}+bx+c& =\hfill & 0\phantom{\rule{2em}{0ex}}a\ne 0\hfill \\ \text{and solve it for}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\text{by completing the square.}\hfill & & & & & \\ \text{Isolate the variable terms on one side.}\hfill & & & \hfill a{x}^{2}+bx& =\hfill & \text{−}c\hfill \\ \\ \text{Make leading coefficient 1, by dividing by a.}\hfill & & & \hfill \frac{a{x}^{2}}{a}+\frac{b}{a}x& =\hfill & -\frac{c}{a}\hfill \\ \\ \text{Simplify.}\hfill & & & \hfill {x}^{2}+\frac{b}{a}x& =\hfill & -\frac{c}{a}\hfill \\ \text{To complete the square, find}\phantom{\rule{0.2em}{0ex}}{\left(\frac{1}{2}·\frac{b}{a}\right)}^{2}\phantom{\rule{0.2em}{0ex}}\text{and add it to both}\hfill & & & \\ \text{sides of the equation.}\phantom{\rule{0.2em}{0ex}}{\left(\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}\hfill & & & \hfill {x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}& =\hfill & -\frac{c}{a}+\frac{{b}^{2}}{4{a}^{2}}\hfill \\ \\ \text{The left side is a perfect square, factor it.}\hfill & & & \hfill {\left(x+\frac{b}{2a}\right)}^{2}& =\hfill & -\frac{c}{a}+\frac{{b}^{2}}{4{a}^{2}}\hfill \\ \\ \begin{array}{c}\text{Find the common denominator of the right side and write}\hfill \\ \text{equivalent fractions with the common denominator.}\hfill \end{array}\hfill & & & \hfill {\left(x+\frac{b}{2a}\right)}^{2}& =\hfill & \frac{{b}^{2}}{4{a}^{2}}-\frac{c·4a}{a·4a}\hfill \\ \\ \text{Simplify.}\hfill & & & \hfill {\left(x+\frac{b}{2a}\right)}^{2}& =\hfill & \frac{{b}^{2}}{4{a}^{2}}-\frac{4ac}{4{a}^{2}}\hfill \\ \\ \text{Combine to one fraction.}\hfill & & & \hfill {\left(x+\frac{b}{2a}\right)}^{2}& =\hfill & \frac{{b}^{2}-4ac}{4{a}^{2}}\hfill \\ \\ \text{Use the square root property.}\hfill & & & \hfill x+\frac{b}{2a}& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ \\ \text{Simplify.}\hfill & & & \hfill x+\frac{b}{2a}& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \\ \text{Add}\phantom{\rule{0.2em}{0ex}}-\frac{b}{2a}\phantom{\rule{0.2em}{0ex}}\text{to both sides of the equation.}\hfill & & & \hfill x& =\hfill & -\frac{b}{2a}±\frac{\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ \\ \text{Combine the terms on the right side.}\hfill & & & \hfill x& =\hfill & \frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}\)

This last equation is the Quadratic Formula.

To use the Quadratic Formula, we substitute the values of \(a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c\) into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Example: How to Solve a Quadratic Equation Using the Quadratic Formula

Solve \(2{x}^{2}+9x-5=0\) by using the Quadratic Formula.

Solution