Learning Objectives

In this section you will:

5.6.1 – Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by $\,-1\,$ and obtain a positive a. Given $\,a{x}^{2}+bx+c=0,$ $a\ne 0,$ we will complete the square as follows:

1. First, move the constant term to the right side of the equal sign:

$a{x}^{2}+bx=-c$
2. As we want the leading coefficient to equal 1, divide through by a:

${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
3. Then, find $\,\frac{1}{2}\,$ of the middle term, and add $\,{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}\,$ to both sides of the equal sign:

${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
5. Now, use the square root property, which gives

$\begin{array}{ccc}\hfill x+\frac{b}{2a}& =& \pm\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ \hfill x+\frac{b}{2a}& =& \frac{\pm\sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
6. Finally, add $\,-\frac{b}{2a}\,$ to both sides of the equation and combine the terms on the right side. Thus,

$x=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}$

Written in standard form, $\,a{x}^{2}+bx+c=0,$ any quadratic equation can be solved using the quadratic formula:

$x=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}$

where a, b, and c are real numbers and $\,a\ne 0.$

How To

1. Make sure the equation is in standard form: $\,a{x}^{2}+bx+c=0.$
2. Make note of the values of the coefficients and constant term, $\,a,b,$ and $\,c.$
3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
4. Calculate and solve.

Solve the quadratic equation: $\,{x}^{2}+5x+1=0.$

Identify the coefficients: $\,a=1,b=5,c=1.\,$ Then use the quadratic formula.

$\begin{array}{ccc}\hfill x& =& \hfill \frac{-\left(5\right)\pm\sqrt{{\left(5\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ & =& \frac{-5\pm\sqrt{25-4}}{2}\hfill \\ & =& \frac{-5\pm\sqrt{21}}{2}\hfill \end{array}$

Use the quadratic formula to solve $\,{x}^{2}+x+2=0.$

First, we identify the coefficients: $\,a=1,b=1,$ and $\,c=2.$

Substitute these values into the quadratic formula.

$\begin{array}{ccc}\hfill x& =& \frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ & =& \frac{-\left(1\right)\pm\sqrt{{\left(1\right)}^{2}-\left(4\right)\cdot \left(1\right)\cdot \left(2\right)}}{2\cdot 1}\hfill \\ & =& \frac{-1\pm\sqrt{1-8}}{2}\hfill \\ & =& \frac{-1\pm\sqrt{-7}}{2}\hfill \\ & =& \frac{-1\pm i\sqrt{7}}{2}\hfill \end{array}$

The solutions to the equation are $\,\frac{-1+i\sqrt{7}}{2}\,$ and $\,\frac{-1-i\sqrt{7}}{2}\,$ or $\,\frac{-1}{2}+\frac{i\sqrt{7}}{2}\,$ and $\,\frac{-1}{2}-\frac{i\sqrt{7}}{2}.$

Try It

Solve the quadratic equation using the quadratic formula: $\,9{x}^{2}+3x-2=0.$

$x=-\frac{2}{3}$

$x=\frac{1}{3}$

Try It

Solve the quadratic equation using the quadratic formula: $\,2{x}^{2}-5x+3=0.$

$x=1$

$x=\frac{3}{2}$

Try It

Solve the quadratic equation using the quadratic formula: $\,\frac{1}{2}{x}^{2}-5x+3=0.$

$x=5 - \sqrt{19}$

$x=5 + \sqrt{19}$

Access these online resources for additional instruction and practice with quadratic equations.

Key Equations

 quadratic formula $x=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a}$

Key Concepts

• A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See (Example 1) and see (Example 2)