Let's Learn Solving Quadratic Equations by the Square Root Method

Section 12.4 Solving Quadratic Equations by the Square Root Method

If \(k\) is a positive number and \(u^2=k\text{,}\) then:

\begin{equation*} u=\sqrt{k}\text{ or } u=-\sqrt{k}\text{.} \end{equation*}

The latter set of equations is frequently abbreviated as

\begin{equation*} u= \pm \sqrt{k} \end{equation*}

which is read aloud as "\(u\) is equal to plus or minus the square root of \(k\text{.}\)"

For example, if \(x^2=64\text{,}\) then \(x= \pm \sqrt{64}\) which, of course, simplifies to \(x= \pm 8\text{.}\) So the original equation has two solutions: \(-8\) and \(8\text{.}\)

Frequently there are additional steps that need to be applied to completely isolate the variable.

Example 12.4.1.

Determine the solutions and the solution set to the equation \((3x+5)^2=16\text{.}\)

Solution

\begin{align*} (3x+5)^2\amp=16\\ 3x+5\amp=\pm \sqrt{16}\\ 3x+5\amp=\pm 4 \end{align*}

When the square root term simplifies to an integer, we split the equation into two separate equations, one using the positive square root and one using the negative square root.

\begin{align*} 3x+5\amp=-4\amp\amp\text{or}\amp 3x+5\amp=4\\ 3x+5\subtractright{5}\amp=-4\subtractright{5}\amp\amp\text{or}\amp 3x+5\subtractright{5}\amp=4\subtractright{5}\\ 3x\amp=-9\amp\amp\text{or}\amp 3x\amp=-1\\ \divideunder{3x}{3}\amp=\divideunder{-9}{3}\amp\amp\text{or}\amp \divideunder{3x}{3}\amp=\divideunder{-1}{3}\\ x\amp=-3\amp\amp\text{or}\amp x\amp=-\frac{1}{3} \end{align*}

The solutions are \(-3\) and \(-\frac{1}{3}\text{.}\) The solution set is \(\left\{-3, -\frac{1}{3}\right\}\text{.}\)

Example 12.4.2.

Determine the solutions and solution set for the equation \(\left(-\frac{3}{2}w+7\right)^2=12\text{.}\)

Solution

We being by applying the square root method.

\begin{align*} \left(-\frac{3}{2}w+7\right)^2\amp=12\\ -\frac{3}{2}w+7\amp=\pm\sqrt{12} \end{align*}

Because \(\sqrt{12}\) is not an integer, there is no reason to split the equation into two equations. We do need to simplify \(\sqrt{12}\text{.}\) Let's do that and also isolate \(w\text{.}\)

\begin{align*} -\frac{3}{2}w+7\amp=\pm\sqrt{12}\\ -\frac{3}{2}w+7\amp=\pm 2\sqrt{3}\\ -\frac{3}{2}w+7\subtractright{7}\amp=\pm 2\sqrt{3}\subtractright{7} \end{align*}

It is customary to have the square root term at the back end of the expression.

\begin{align*} -\frac{3}{2}w\amp=-7\pm 2\sqrt{3}\\ \multiplyleft{-\frac{2}{3}}-\frac{3}{2}w\amp=\multiplyleft{-\frac{2}{3}}\left(-7\pm 2\sqrt{3}\right)\\ w\amp=-\frac{2}{3} \cdot -\frac{7}{1} \pm -\frac{2}{3} \cdot \frac{2\sqrt{3}}{1}\\ w\amp=-\frac{14}{3} \pm \frac{4\sqrt{3}}{3} \end{align*}

Technically the expression after the plus/minus sign should be negative. However, because the term is being both added and subtracted, we can achieve the same effect by adding and subtracting the positive expression. In that sense the negative sign is redundant and is always omitted in this situation. It is customary to go ahead and write the two terms over one denominator.

\begin{align*} w\amp=-\frac{14}{3} \pm \frac{4\sqrt{3}}{3}\\ w\amp=\frac{-14\pm 4\sqrt{3}}{3} \end{align*}

When stating the solutions and solution set, we detach the solution with the addition sign from the solution with the subtraction sign.

The solutions are \(\frac{-14-4\sqrt{3}}{3}\) and \(\frac{-14+4\sqrt{3}}{3}\text{.}\)

The solution set is \(\left\{\frac{-14-4\sqrt{3}}{3},\frac{-14+4\sqrt{3}}{3}\right\}\text{.}\)

Example 12.4.3.

Determine the solutions and solution set to the equation \((5x-2)^2=-25\text{.}\)

Solution

There is no real number that squares to \(-25\text{.}\) Unless otherwise stated, the assumption is that we are solving the equation over the real numbers. As such, the stated equation has no solutions (over the real numbers) and the solution set is \(\emptyset\) (the empty set).

Example 12.4.4.

Determine the solutions and solution set to the equation \(3(5-6y)^2=25\text{.}\)

Solution

Before applying the square root method, we need to isolate the square root expression by dividing both sides of the equation by \(3\text{.}\)

\begin{align*} 3(5-6y)^2\amp=25\\ \divideunder{(5-6y)^2}{3}\amp=\divideunder{25}{3}\\ (5-6y)^2\amp=\frac{25}{3}\\ 5-6y\amp=\pm\sqrt{\frac{25}{3}} \end{align*}

We need to simplify the square root expression. This includes rationalizing the denominator. Let's do that and continue to isolate \(y\text{.}\)

\begin{align*} 5-6y\amp=\pm\sqrt{\frac{25}{3}}\\ 5-6y\amp=\pm\frac{\sqrt{25}}{\sqrt{3}}\\ 5-6y\amp=\pm\frac{5}{\sqrt{3}}\\ 5-6y\amp=\pm\frac{5}{\sqrt{3}}\multiplyright{\frac{\sqrt{3}}{\sqrt{3}}}\\ 5-6y\amp=\pm\frac{5\sqrt{3}}{3}\\ 5-6y\subtractright{5}\amp=\pm\frac{5\sqrt{3}}{3}\subtractright{5} \end{align*}

As in a previous example, we want the plus/minus expression at the rear. We also need to establish a common denominator on the right side of the equation and combine those two terms.

\begin{align*} -6y\amp=-5\pm\frac{5\sqrt{3}}{3}\\ -6y\amp=-\frac{15}{3}\pm\frac{5\sqrt{3}}{3}\\ -6y\amp=\frac{-15\pm 5\sqrt{3}}{3} \end{align*}

We're going to finish up by multiplying both sides of the equation by \(-\frac{1}{6}\text{.}\) We'll distribute the negative sign through the numerator on the right side of the equation. However, as discussed in a previous example, there's no need to make the expression after the plus/minus sign negative.

\begin{align*} \multiplyleft{-\frac{1}{6}}(-6y)\amp=\multiplyleft{-\frac{1}{6}}\frac{-15\pm 5\sqrt{3}}{3}\\ y\amp=\frac{15\pm 5\sqrt{3}}{18} \end{align*}

The solutions are \(\frac{15-5\sqrt{3}}{18}\) and \(\frac{15+5\sqrt{3}}{18}\text{.}\)

The solution set is \(\left\{\frac{15-5\sqrt{3}}{18},\frac{15+5\sqrt{3}}{18}\right\}\text{.}\)

Example 12.4.5.

Determine the solutions and the solutions set for the equation \(7-9(8-t)^2=7\text{.}\)

Solution

We begin by isolating the square root expression.

\begin{align*} 7-9(8-t)^2\amp=7\\ 7-9(8-t)^2\subtractright{7}\amp=7\subtractright{7}\\ -9(8-t)^2\amp=0\\ \divideunder{-9(8-t)^2}{-9}\amp=\divideunder{0}{-9}\\ (8-t)^2\amp=0\\ 8-t\amp=\pm\sqrt{0} \end{align*}

Because there is only one square root of zero (that being zero), this equation has only one solution.

\begin{align*} 8-t\amp=0\\ 8-t\subtractright{8}\amp=0\subtractright{8}\\ -t\amp=-8\\ \divideunder{-t}{-1}\amp=\divideunder{-8}{-1}\\ t\amp=8 \end{align*}

The only solution is \(8\text{.}\)

The solution set is \(\{8\}\text{.}\)

Exercises Exercises

Use the square root method to solve each of the following quadratic equations over the real numbers. Make sure that all solutions are completely simplified. State the solutions to each equation as well as the solution set to each equation.

1.

Determine the solutions and the solution set to the equation \((2t-1)^2=9\text{.}\)

Solution

We begin by applying the square root property.

\begin{align*} (2t-1)^2\amp=9\\ 2t-1\amp=\pm \sqrt{9}\\ 2t-1\amp=\pm 3 \end{align*}

Because the square root expression simplified to an iteger. We split the equation into two separate equations.

\begin{align*} 2t-1\amp =-3\amp\amp\text{ or }\amp 2t-1\amp=3\\ 2t-1\addright{1}\amp =-3\addright{1}\amp\amp\text{or}\amp 2t-1\addright{1}\amp=3\addright{1}\\ 2t\amp =-2\amp\amp\text{or}\amp 2t\amp=4\\ \divideunder{2t}{2}\amp =\divideunder{-2}{2}\amp\amp\text{or}\amp \divideunder{2t}{2}\amp=\divideunder{4}{2}\\ t\amp =-1\amp\amp\text{or}\amp t\amp=2 \end{align*}

The solutions are \(-1\) and \(2\text{.}\)

The solution set is \(\{-1, 2\}\text{.}\)

2.

Determine the solutions and the solution set to the equation \((3x-2)^2=72\text{.}\)

Solution

We begin by applying the square root property.

\begin{align*} (3x-2)^2\amp=72\\ 3x-2\amp=\pm \sqrt{72}\\ 3x-2\amp=\pm \sqrt{36 \cdot 2}\\ 3x-2\amp=\pm 6\sqrt{2}\\ 3x-2\addright{2}\amp=\pm 6\sqrt{2}\addright{2}\\ 3x\amp=2\pm 6\sqrt{2}\\ \multiplyleft{\frac{1}{3}}3x\amp=\multiplyleft{\frac{1}{3}}(2\pm 6\sqrt{2})\\ x\amp=\frac{2\pm 6\sqrt{2}}{3} \end{align*}

The solutions are \(\frac{2-6\sqrt{2}}{3}\) and \(\frac{2+6\sqrt{2}}{3}\text{.}\)

The solution set is \(\left\{\frac{2-6\sqrt{2}}{3}, \frac{2+6\sqrt{2}}{3}\right\}\text{.}\)

3.

Determine the solution and solution set to the equation \(5-3(x+8)^2=17\text{.}\)

Solution

We begin by isolating the square root expression.

\begin{align*} 5-3(x+8)^2\amp=17\\ 5-3(x+8)^2\subtractright{5}\amp=17\subtractright{5}\\ -3(x+8)^2\amp=12\\ \divideunder{-3(x+8)^2}{-3}\amp=\divideunder{12}{-3}\\ (x+8)^2\amp=-4 \end{align*}

There is no real number that squares to \(-4\text{.}\) As such, the stated equation has no solutions over the real numbers and its solution set is \(\emptyset\text{.}\)

4.

Determine the solution and solution set to the equation \(5(2t-3)^2-9=15\text{.}\)

Solution

We begin by isolating the square root expression.

\begin{align*} 5(2t-3)^2-9\amp=15\\ 5(2t-3)^2-9\addright{9}\amp=15\addright{9}\\ 5(2t-3)^2\amp=24\\ \divideunder{5(2t-3)^2}{5}\amp=\divideunder{24}{5}\\ (2t-3)^2\amp=\frac{24}{5} \end{align*}

We can now apply the square root method and then isolate \(t\text{.}\) We need to be mindful to completely simplify the square root expression and to establish a common denominator in our final expression.

\begin{align*} (2t-3)^2\amp=\frac{24}{5}\\ 2t-3\amp=\pm\sqrt{\frac{24}{5}}\\ 2t-3\amp=\pm\frac{\sqrt{24}}{\sqrt{5}}\\ 2t-3\amp=\frac{2\sqrt{6}}{\sqrt{5}}\\ 2t-3\amp=\frac{2\sqrt{6}}{\sqrt{5}}\multiplyright{\frac{\sqrt{5}}{\sqrt{5}}}\\ 2t-3\amp=\pm\frac{2\sqrt{30}}{5}\\ 2t-3\addright{3}\amp=\pm\frac{2\sqrt{30}}{5}\addright{3}\\ 2t\amp=3\pm\frac{2\sqrt{30}}{5}\\ 2t\amp=\frac{15}{3}\pm\frac{2\sqrt{30}}{5}\\ 2t\amp=\frac{15\pm 2\sqrt{30}}{5}\\ \multiplyleft{\frac{1}{2}}2t\amp=\multiplyleft{\frac{1}{2}}\frac{15\pm 2\sqrt{30}}{5}\\ y\amp=\frac{15 \pm 2\sqrt{30}}{10} \end{align*}

The solutions are \(\frac{15-2\sqrt{30}}{10}\) and \(\frac{15+2\sqrt{30}}{10}\text{.}\)

The solutions set is \(\left\{\frac{15-2\sqrt{30}}{10},\frac{15+2\sqrt{30}}{10}\right\}\text{.}\)

5.

Determine the solution and solution set to the equation \(\frac{(8-y)^2}{3}=6\)

Solution

We begin by isolating the square root expression.

\begin{align*} \frac{(8-y)^2}{3}\amp=6\\ \multiplyleft{3}\frac{(8-y)^2}{3}\amp=\multiplyleft{3}6\\ (8-y)^2\amp=18 \end{align*}

We can now apply the square root method. We will then simplify the square root expression and work towards isolating \(y\text{.}\)

\begin{align*} (8-y)^2\amp=18\\ 8-y\amp=\pm\sqrt{18}\\ 8-y\amp=\pm 3\sqrt{2}\\ 8-y\subtractright{8}\amp=\pm 3\sqrt{2}\subtractright{8}\\ -y\amp=-8\pm 3\sqrt{2} \end{align*}

At this point we can isolation \(y\) by multiplying both sides of the equation by \(-1\text{.}\) Technically we need to distribute the negative sign to the two terms on the right side of the equation. But as previously discussed, having a negative sign after a plus/minus sign is redundant and, hence, not done.

\begin{align*} -y\amp=-8\pm 3\sqrt{2}\\ \multiplyleft{-1}-y\amp=\multiplyleft{-1}\left(-8\pm 3\sqrt{2}\right)\\ y\amp=8\pm 3\sqrt{2} \end{align*}

The solutions are \(8-3\sqrt{2}\) and \(8+3\sqrt{2}\text{.}\)

The solution set is \(\left\{8-3\sqrt{2},8+3\sqrt{2}\right\}\text{.}\)

6.

Determine the solution and solution set to the equation \(2(3w+8)^2+1=33\text{.}\)

Solution

We begin by isolating the square root expression.

\begin{align*} 2(3w+8)^2+1\amp=33\\ 2(3w+8)^2+1\subtractright{1}\amp=33\subtractright{1}\\ 2(3w+8)^2\amp=32\\ \divideunder{2(3w+8)^2}{2}\amp=\divideunder{32}{2}\\ (3w+8)^2\amp=16 \end{align*}

We can now apply the square root method.

\begin{align*} (3w+8)^2\amp=16\\ 3w+8\amp=\pm\sqrt{16}\\ 3w+8\amp=\pm 4 \end{align*}

Because the square root expression simplified to an integer, we split the equation into two separate equations.

\begin{align*} 3w+8\amp=-4\amp\amp\text{or}\amp 3w+8\amp=4\\ 3w+8\subtractright{8}\amp=-4\subtractright{8}\amp\amp\text{or}\amp 3w+8\subtractright{8}\amp=4\subtractright{8}\\ 3w\amp=-12\amp\amp\text{or}\amp 3w\amp=-4\\ \divideunder{3w}{3}\amp=\divideunder{-12}{3}\amp\amp\text{or}\amp \divideunder{3w}{3}\amp=\divideunder{-4}{3}\\ w\amp-4\amp\amp\text{or}\amp w\amp=-\frac{4}{3} \end{align*}

The solutions are \(-4\) and \(-\frac{4}{3}\text{.}\)

The solution set is \(\left\{-4,-\frac{4}{3}\right\}\text{.}\)