Solving Quadratic Equations of the Form <em>ax</em><sup>2</sup> = <em>k</em>
Solve Quadratic Equations of the Form ax2 = k Using the Square Root Property
We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation \({x}^{2}=9\).
\(\begin{array}{cccccc}& & & \hfill {x}^{2}& =\hfill & \hfill 9\\ \text{Put the equation in standard form.}\hfill & & & \hfill {x}^{2}-9& =\hfill & \hfill 0\\ \text{Factor the left side.}\hfill & & & \hfill \left(x-3\right)\left(x+3\right)& =\hfill & \hfill 0\\ \text{Use the Zero Product Property.}\hfill & & & \hfill \left(x-3\right)=0,\phantom{\rule{0.5em}{0ex}}\left(x+3\right)& =\hfill & \hfill 0\\ \text{Solve each equation.}\hfill & & & \hfill x=3,\phantom{\rule{2.8em}{0ex}}x& =\hfill & \hfill -3\\ \text{Combine the two solutions into}\phantom{\rule{0.2em}{0ex}}±\phantom{\rule{0.2em}{0ex}}\text{form.}\hfill & & & \hfill x& =\hfill & \hfill ±\phantom{\rule{0.2em}{0ex}}3\\ \text{(The solution is read}\phantom{\rule{0.2em}{0ex}}‘x\phantom{\rule{0.2em}{0ex}}\text{is equal to positive or negative 3.’)}\hfill & & \end{array}\)
We can easily use factoring to find the solutions of similar equations, like \({x}^{2}=16\) and \({x}^{2}=25\), because 16 and 25 are perfect squares. But what happens when we have an equation like \({x}^{2}=7\)? Since 7 is not a perfect square, we cannot solve the equation by factoring.
These equations are all of the form \({x}^{2}=k\).
We defined the square root of a number in this way:
\(\text{If}\phantom{\rule{0.2em}{0ex}}{n}^{2}=m,\phantom{\rule{0.2em}{0ex}}\text{then}\phantom{\rule{0.2em}{0ex}}n\phantom{\rule{0.2em}{0ex}}\text{is a square root of}\phantom{\rule{0.2em}{0ex}}m.\)
This leads to the Square Root Property.
Square Root Property
If \({x}^{2}=k\), and \(k\ge 0\), then \(x=\sqrt{k}\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=\text{−}\sqrt{k}\).
Notice that the Square Root Property gives two solutions to an equation of the form \({x}^{2}=k\): the principal square root of \(k\) and its opposite. We could also write the solution as \(x=±\phantom{\rule{0.2em}{0ex}}\sqrt{k}\).
Now, we will solve the equation \({x}^{2}=9\) again, this time using the Square Root Property.
\(\begin{array}{cccccc}& & & \hfill {x}^{2}& =\hfill & 9\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill x& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{9}\hfill \\ \text{Simplify the radical.}\hfill & & & \hfill x& =\hfill & ±\phantom{\rule{0.2em}{0ex}}3\hfill \\ \text{Rewrite to show the two solutions.}\hfill & & & \hfill x=3,x& =\hfill & -3\hfill \end{array}\)
What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation \({x}^{2}=7\).
\(\begin{array}{cccc}\begin{array}{}\\ \\ \text{Use the Square Root Property.}\hfill \end{array}\hfill & & & \hfill \begin{array}{ccc}\hfill {x}^{2}& \hfill =\hfill & 7\hfill \\ \hfill x& \hfill =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{7}\hfill \end{array}\hfill \\ \text{Rewrite to show two solutions.}\hfill & & & \hfill x=\sqrt{7},\phantom{\rule{1em}{0ex}}x=\text{−}\sqrt{7}\hfill \\ \text{We cannot simplify}\phantom{\rule{0.2em}{0ex}}\sqrt{7},\phantom{\rule{0.2em}{0ex}}\text{so we leave the answer as a radical.}\hfill & & \end{array}\)
Example
Solve: \({x}^{2}=169\).
Solution
\(\begin{array}{cccc}\begin{array}{}\\ \\ \text{Use the Square Root Property.}\hfill \\ \text{Simplify the radical.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{ccc}\hfill {x}^{2}& =\hfill & 169\hfill \\ \hfill x& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{169}\hfill \\ \hfill x& =\hfill & ±\phantom{\rule{0.2em}{0ex}}13\hfill \end{array}\hfill \\ \text{Rewrite to show two solutions.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}x=13,\phantom{\rule{1em}{0ex}}x=-13\hfill \end{array}\)
Example: How to Solve a Quadratic Equation of the Form \({ax}^{2}=k\) Using the Square Root Property
Solve: \({x}^{2}-48=0\).
Solution
Solve a quadratic equation using the Square Root Property.
- Isolate the quadratic term and make its coefficient one.
- Use Square Root Property.
- Simplify the radical.
- Check the solutions.
To use the Square Root Property, the coefficient of the variable term must equal 1. In the next example, we must divide both sides of the equation by 5 before using the Square Root Property.
Example
Solve: \(5{m}^{2}=80\).
Solution
| The quadratic term is isolated. | \(\phantom{\rule{0.1em}{0ex}}5{m}^{2}=80\) | |
| Divide by 5 to make its cofficient 1. | \(\frac{5{m}^{2}}{5}=\frac{80}{5}\) | |
| Simplify. | \(\phantom{\rule{0.6em}{0ex}}{m}^{2}=16\) | |
| Use the Square Root Property. | \(\phantom{\rule{1.1em}{0ex}}m=±\phantom{\rule{0.2em}{0ex}}\sqrt{16}\) | |
| Simplify the radical. | \(\phantom{\rule{1.1em}{0ex}}m=±\phantom{\rule{0.2em}{0ex}}4\) | |
| Rewrite to show two solutions. | \(m=4,m=\text{−}4\) | |
| Check the solutions. |
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The Square Root Property started by stating, ‘If \({x}^{2}=k\), and \(k\ge 0\)’. What will happen if \(k<0\)? This will be the case in the next example.
Example
Solve: \({q}^{2}+24=0\).
Solution
\(\begin{array}{cccc}\begin{array}{}\\ \\ \text{Isolate the quadratic term.}\hfill \\ \text{Use the Square Root Property.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{ccc}\hfill {q}^{2}+24& =\hfill & 0\hfill \\ \hfill {q}^{2}& =\hfill & -24\hfill \\ \hfill q& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{-24}\hfill \end{array}\\ \text{The}\phantom{\rule{0.2em}{0ex}}\sqrt{-24}\phantom{\rule{0.2em}{0ex}}\text{is not a real number.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{There is no real solution.}\hfill \end{array}\)
Remember, we first isolate the quadratic term and then make the coefficient equal to one.
Example
Solve: \(\frac{2}{3}{u}^{2}+5=17\).
Solution
| \(\frac{2}{3}{u}^{2}+5=17\) | ||
| Isolate the quadratic term. | \(\phantom{\rule{1.7em}{0ex}}\frac{2}{3}{u}^{2}=12\) | |
| Multiply by \(\frac{3}{2}\) to make the coefficient 1. | \(\phantom{\rule{0.5em}{0ex}}\frac{3}{2}·\frac{2}{3}{u}^{2}=\frac{3}{2}·12\) | |
| Simplify. | \(\phantom{\rule{2.2em}{0ex}}{u}^{2}=18\) | |
| Use the Square Root Property. | \(\phantom{\rule{2.7em}{0ex}}u=±\phantom{\rule{0.2em}{0ex}}\sqrt{18}\) | |
| Simplify the radical. | \(\phantom{\rule{2.7em}{0ex}}u=±\phantom{\rule{0.2em}{0ex}}\sqrt{9}\sqrt{2}\) | |
| Simplify. | \(\phantom{\rule{2.7em}{0ex}}u=±\phantom{\rule{0.2em}{0ex}}3\sqrt{2}\) | |
| Rewrite to show two solutions. | \(u=3\sqrt{2},u=\text{−}3\sqrt{2}\) | |
| Check. |
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The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.
Example
Solve: \(2{c}^{2}-4=45\).
Solution
\(\begin{array}{c}\begin{array}{cccccc}& & & \hfill \phantom{\rule{4em}{0ex}}2{c}^{2}-4& =\hfill & 45\hfill \\ \text{Isolate the quadratic term.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}2{c}^{2}& =\hfill & 49\hfill \\ \text{Divide by 2 to make the coefficient 1.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\frac{2{c}^{2}}{2}& =\hfill & \frac{49}{2}\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{c}^{2}& =\hfill & \frac{49}{2}\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}c& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\sqrt{\frac{49}{2}}\hfill \\ \text{Simplify the radical.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}c& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{49}}{\sqrt{2}}\hfill \\ \text{Rationalize the denominator.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}c& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\frac{\sqrt{49}·\sqrt{2}}{\sqrt{2}·\sqrt{2}}\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}c& =\hfill & ±\phantom{\rule{0.2em}{0ex}}\frac{7\sqrt{2}}{2}\hfill \end{array}\hfill \\ \\ \text{Rewrite to show two solutions.}\phantom{\rule{9em}{0ex}}c=\frac{7\sqrt{2}}{2},\phantom{\rule{1em}{0ex}}c=-\frac{7\sqrt{2}}{2}\hfill & & \\ \text{Check. We leave the check for you.}\hfill & & \end{array}\)
This lesson is part of:
Introducing Quadratic Equations