Solve Quadratic Equations Using The Square Root Property

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Solve Quadratic Equations of the form $a{x}^{2}=k$ 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 x2 = 9.

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \text{Put the equation in standard form.}\hfill \\ \text{Factor the difference of squares.}\hfill \end{array}\hfill & & & \hfill \begin{array}{ccc}\hfill {x}^{2}& =\hfill & 9\hfill \\ \hfill {x}^{2}-9& =\hfill & 0\hfill \\ \hfill \left(x-3\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the Zero Product Property.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \hfill \begin{array}{cccccccc}\hfill x-3& =\hfill & 0\hfill & & & \hfill x-3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 3\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill \end{array}$

We can easily use factoring to find the solutions of similar equations, like x2 = 16 and x2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, and

But what happens when we have an equation like x2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

$\text{If}\phantom{\rule{0.2em}{0ex}}{n}^{2}=m,\phantom{\rule{0.2em}{0ex}}\text{then}\phantom{\rule{0.2em}{0ex}}n\phantom{\rule{0.2em}{0ex}}\text{is a square root of}\phantom{\rule{0.2em}{0ex}}m.$

Since these equations are all of the form x2 = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Square Root Property

If x2 = k, then

$x=\sqrt{k}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=\text{−}\sqrt{k}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=±\sqrt{k}.$

Notice that the Square Root Property gives two solutions to an equation of the form x2 = k, the principal square root of and its opposite. We could also write the solution as $x=±\sqrt{k}.$ We read this as x equals positive or negative the square root of k.

Now we will solve the equation x2 = 9 again, this time using the Square Root Property.

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \text{Use the Square Root Property.}\hfill \\ \\ \\ \end{array}\hfill & & & \hfill \begin{array}{ccc}\hfill {x}^{2}& =\hfill & 9\hfill \\ \hfill x& =\hfill & ±\sqrt{9}\hfill \\ \hfill x& =\hfill & ±3\hfill \end{array}\hfill \\ & & & \hfill \text{So}\phantom{\rule{0.2em}{0ex}}x=3\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}x=-3.\hfill \end{array}$

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2 = 7.

$\begin{array}{cccc}& & & \hfill {x}^{2}=7\hfill \\ \text{Use the Square Root Property.}\hfill & & & \hfill x=\sqrt{7},\phantom{\rule{1em}{0ex}}x=\text{−}\sqrt{7}\hfill \end{array}$

We cannot simplify $\sqrt{7}$ , so we leave the answer as a radical.

How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property

Solve: ${x}^{2}-50=0.$

Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.

Step two is to use Square Root Property. Remember to write the plus or minus symbol. The equation created is x equals the positive or negative square root of 50.

Step three is to simplify the radical if possible. Continue to write equivalent equations. X equals the positive or negative square root of twenty five times the square root of five. X equals positive or negative five times the square root of five. Rewrite to show two solutions: x equals five times the square root of five or x equals negative five times the square root of five.

Step four is to check the solutions. Substitute x equals five times the square root of five and or x equals negative five times the square root of five into the original equation. Start with x equals five times the square root of five. x squared minus fifty equals zero. Does the square of five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero. Next check x equals negative five times the square root of five. x squared minus fifty equals zero. Does the square of negative five times the square root of five minus fifty equal zero? Does twenty five times two minus fifty equal zero? Yes, because zero equals zero.

Solve: ${x}^{2}-48=0.$

$x=4\sqrt{3},x=-4\sqrt{3}$

Solve: ${y}^{2}-27=0.$

$y=3\sqrt{3},y=-3\sqrt{3}$

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

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.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Solve: $3{z}^{2}=108.$

$3{z}^{2}=108$
The quadratic term is isolated. Divide by 3 to make its coefficient 1.	$\frac{3{z}^{2}}{3}=\frac{108}{3}$
Simplify.	${z}^{2}=36$
Use the Square Root Property.	$\phantom{\rule{2em}{0ex}}z=±\sqrt{36}$
Simplify the radical.	$\phantom{\rule{1.1em}{0ex}}z=±6$
Rewrite to show two solutions.	$z=6,\phantom{\rule{0.5em}{0ex}}z=-6$
Check the solutions:

Solve: $2{x}^{2}=98.$

Solve: $5{m}^{2}=80.$

The Square Root Property states ‘If ${x}^{2}=k$ ,’ What will happen if This will be the case in the next example.

Solve: ${x}^{2}+72=0$ .

${x}^{2}+72=0$
Isolate the quadratic term.	$\phantom{\rule{2.15em}{0ex}}{x}^{2}=-72$
Use the Square Root Property.	$\phantom{\rule{2.6em}{0ex}}x=±\sqrt{-72}$
Simplify using complex numbers.	$\phantom{\rule{2.6em}{0ex}}x=±\sqrt{72}\phantom{\rule{0.2em}{0ex}}i$
Simplify the radical.	$\phantom{\rule{2.6em}{0ex}}x=±6\sqrt{2}\phantom{\rule{0.2em}{0ex}}i$
Rewrite to show two solutions.	$x=6\sqrt{2}\phantom{\rule{0.2em}{0ex}}i,\phantom{\rule{0.5em}{0ex}}x=-6\sqrt{2}\phantom{\rule{0.2em}{0ex}}i$
Check the solutions:

Solve: ${c}^{2}+12=0.$

$c=2\sqrt{3}i,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}c=-2\sqrt{3}i$

Solve: ${q}^{2}+24=0.$

$c=2\sqrt{6}i,\phantom{\rule{0.2em}{0ex}}\text{}\phantom{\rule{0.2em}{0ex}}c=-2\sqrt{6}i$

Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Solve: $\frac{2}{3}{u}^{2}+5=17.$

$\frac{2}{3}{u}^{2}+5=17$
Isolate the quadratic term.
Multiply by $\frac{3}{2}$ to make the coefficient 1.
Simplify.
Use the Square Root Property.
Simplify the radical.
Simplify.
Rewrite to show two solutions.
Check:

Solve: $\frac{1}{2}{x}^{2}+4=24.$

$x=2\sqrt{10},\phantom{\rule{0.5em}{0ex}}x=-2\sqrt{10}$

Solve: $\frac{3}{4}{y}^{2}-3=18.$

$y=2\sqrt{7},\phantom{\rule{0.5em}{0ex}}y=-2\sqrt{7}$

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Solve: $2{x}^{2}-8=41.$


Isolate the quadratic term.
Divide by to make the coefficient 1.
Simplify.
Use the Square Root Property.
Rewrite the radical as a fraction of square roots.
Rationalize the denominator.
Simplify.
Rewrite to show two solutions.
Check: We leave the check for you.