5.7 Solving Equations with Square Roots

Solving Equations with Square Roots

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Solve Radical Equations

In this section we will solve equations that have the variable in the radicand of a square root. Equations of this type are called radical equations.

As usual, in solving these equations, what we do to one side of an equation we must do to the other side as well. Since squaring a quantity and taking a square root are ‘opposite’ operations, we will square both sides in order to remove the radical sign and solve for the variable inside.

But remember that when we write $\sqrt{a}$ we mean the principal square root. So $the square root of a greater than or equal to 0$ always. When we solve radical equations by squaring both sides we may get an algebraic solution that would make $\sqrt{a}$ negative. This algebraic solution would not be a solution to the original radical equation; it is an extraneous solution.

Example 1

For the equation $\sqrt{x + 2} = x$ :

ⓐ Is $x equals 2$ a solution? ⓑ Is $x equals −1$ a solution?

Show solution

ⓐ Is $x equals 2$ a solution?


Let x = 2.
Simplify.

	2 is a solution.

ⓑ Is $x equals −1$ a solution?


Let x = −1.
Simplify.

	−1 is not a solution.
	−1 is an extraneous solution to the equation.

Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring.

$For a \geq 0, {(\sqrt{a})}^{2} = a$

When we use a radical sign, we mean the principal, nonnegative root. If an equation has a square root equal to a negative number, that equation will have no solution.

If one side of the equation is a binomial, we use the binomial squares formula when we square it.

Don’t forget the middle term!

Example 6

Solve: $\sqrt{p - 1} + 1 = p$ .

Show solution


To isolate the radical, subtract 1 from both sides.
Simplify.
Square both sides of the equation.
Simplify, then solve the new equation.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the zero product property.
Solve each equation.
Check the answers.

	The solutions are p = 1, p = 2.

Example 7

Solve: $\sqrt{r + 4} - r + 2 = 0$ .

Show solution

	$\sqrt{r + 4} - r + 2 = 0$
Isolate the radical.	$\sqrt{r + 4} = r - 2$
Square both sides of the equation.	${(\sqrt{r + 4})}^{2} = {(r - 2)}^{2}$
Solve the new equation.	$r + 4 = r^{2} - 4 r + 4$
It is a quadratic equation, so get zero on one side.	$0 = r^{2} - 5 r$
Factor the right side.	$0 = r (r - 5)$
Use the zero product property.	$0 = r 0 = r - 5$
Solve the equation.	$r = 0 r = 5$
Check the answer.
	The solution is $r equals 5$ .
	$r equals 0$ is an extraneous solution.

When there is a coefficient in front of the radical, we must square it, too.

Example 8

Solve: $3 \sqrt{3 x - 5} - 8 = 4$ .

Show solution

	$3 \sqrt{3 x - 5} - 8 = 4$
Isolate the radical.	$3 \sqrt{3 x - 5} = 12$
Square both sides of the equation.	${(3 \sqrt{3 x - 5})}^{2} = {(12)}^{2}$
Simplify, then solve the new equation.	$9 (3 x - 5) = 144$
Distribute.	$27 x - 45 = 144$
Solve the equation.	$27 x equals 189$
	$x equals 7$
Check the answer.
	The solution is $x equals 7$ .

Example 9

Solve: $\sqrt{4 z - 3} = \sqrt{3 z + 2}$ .

Show solution

	$\begin{matrix} \sqrt{4 z - 3} = \sqrt{3 z + 2} \end{matrix}$
The radical terms are isolated.	$\begin{matrix} \sqrt{4 z - 3} = \sqrt{3 z + 2} \end{matrix}$
Square both sides of the equation.	$\begin{matrix} {(\sqrt{4 z - 3})}^{2} = {(\sqrt{3 z + 2})}^{2} \end{matrix}$
Simplify, then solve the new equation.	$\begin{matrix} \begin{matrix} 4 z - 3 = 3 z + 2 \\ z - 3 = 2 \\ z = 5 \end{matrix} \end{matrix}$
Check the answer.
We leave it to you to show that 5 checks!	The solution is $z equals 5.$

Sometimes after squaring both sides of an equation, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and square both sides of the equation again.

Example 10

Solve: $\sqrt{m} + 1 = \sqrt{m + 9}$ .

Show solution

	$\sqrt{m} + 1 = \sqrt{m + 9}$
The radical on the right side is isolated. Square both sides.	${(\sqrt{m} + 1)}^{2} = {(\sqrt{m + 9})}^{2}$
Simplify—be very careful as you multiply!	$m + 2 \sqrt{m} + 1 = m + 9$
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical.	$2 \sqrt{m} = 8$
Square both sides.	${(2 \sqrt{m})}^{2} = {(8)}^{2}$
Simplify, then solve the new equation.	$4 m equals 64$
	$m equals 16$
	Check the answer.
We leave it to you to show that $m equals 16$ checks!	The solution is $m equals 16.$

Example 11

Solve: $\sqrt{q - 2} + 3 = \sqrt{4 q + 1}$ .

Show solution

	$\sqrt{q - 2} + 3 = \sqrt{4 q + 1}$
The radical on the right side is isolated. Square both sides.	${(\sqrt{q - 2} + 3)}^{2} = {(\sqrt{4 q + 1})}^{2}$
Simplify.	$q - 2 + 6 \sqrt{q - 2} + 9 = 4 q + 1$
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical.	$6 \sqrt{q - 2} = 3 q - 6$
Square both sides.	${(6 \sqrt{q - 2})}^{2} = {(3 q - 6)}^{2}$
Simplify, then solve the new equation.	$36 (q - 2) = 9 q^{2} - 36 q + 36$
Distribute.	$36 q - 72 = 9 q^{2} - 36 q + 36$
It is a quadratic equation, so get zero on one side.	$0 = 9 q^{2} - 72 q + 108$
Factor the right side.	$0 = 9 (q^{2} - 8 q + 12)$
	$0 = 9 (q - 6) (q - 2)$
Use the zero product property.	$\begin{matrix} q - 6 = 0 & q - 2 = 0 \\ q = 6 & q = 2 \end{matrix}$
The checks are left to you. (Both solutions should work.)	$The solutions are q = 6 and q = 2 .$

Practice Makes Perfect

Solve Radical Equations

In the following exercises, check whether the given values are solutions.

For the equation $\sqrt{x + 12} = x$ : ⓐ Is $x equals 4$ a solution? ⓑ Is $x equals −3$ a solution?

Show solution

ⓐ $yes$ ⓑ $no$

For the equation $\sqrt{− y + 20} = y$ : ⓐ Is $y equals 4$ a solution? ⓑ Is $y equals −5$ a solution?

For the equation $\sqrt{t + 6} = t$ : ⓐ Is $t equals −2$ a solution? ⓑ Is $t equals 3$ a solution?

Show solution

ⓐ no ⓑ yes

For the equation $\sqrt{u + 42} = u$ : ⓐ Is $u equals −6$ a solution? ⓑ Is $u equals 7$ a solution?

In the following exercises, solve.

$\sqrt{5 y + 1} = 4$

Show solution

$\sqrt{7 z + 15} = 6$

$\sqrt{5 x - 6} = 8$

Show solution

$\sqrt{4 x - 3} = 7$

$\sqrt{2 m - 3} - 5 = 0$

Show solution

$\sqrt{2 n - 1} - 3 = 0$

$\sqrt{6 v - 2} - 10 = 0$

Show solution

$17$

$\sqrt{4 u + 2} - 6 = 0$

$\sqrt{5 q + 3} - 4 = 0$

Show solution

$\frac{13}{5}$

$\sqrt{4 m + 2} + 2 = 6$

$\sqrt{6 n + 1} + 4 = 8$

Show solution

$\frac{5}{2}$

$\sqrt{2 u - 3} + 2 = 0$

$\sqrt{5 v - 2} + 5 = 0$

Show solution

$no solution$

$\sqrt{3 z - 5} + 2 = 0$

$\sqrt{2 m + 1} + 4 = 0$

Show solution

$no solution$

ⓐ $\sqrt{u - 3} + 3 = u$
ⓑ $\sqrt{x + 1} - x + 1 = 0$

ⓐ $\sqrt{v - 10} + 10 = v$
ⓑ $\sqrt{y + 4} - y + 2 = 0$

Show solution

ⓐ $10, 11$ ⓑ $5$

ⓐ $\sqrt{r - 1} - r = −1$
ⓑ $\sqrt{z + 100} - z + 10 = 0$

ⓐ $\sqrt{s - 8} - s = −8$
ⓑ $\sqrt{w + 25} - w + 5 = 0$

Show solution

ⓐ $8, 9$ ⓑ $11$

$3 \sqrt{2 x - 3} - 20 = 7$

$2 \sqrt{5 x + 1} - 8 = 0$

Show solution

$3$

$2 \sqrt{8 r + 1} - 8 = 2$

$3 \sqrt{7 y + 1} - 10 = 8$

Show solution

$5$

$\sqrt{3 u - 2} = \sqrt{5 u + 1}$

$\sqrt{4 v + 3} = \sqrt{v - 6}$

Show solution

not a real number

$\sqrt{8 + 2 r} = \sqrt{3 r + 10}$

$\sqrt{12 c + 6} = \sqrt{10 - 4 c}$

Show solution

$\frac{1}{4}$

ⓐ $\sqrt{a} + 2 = \sqrt{a + 4}$
ⓑ $\sqrt{b - 2} + 1 = \sqrt{3 b + 2}$

ⓐ $\sqrt{r} + 6 = \sqrt{r + 8}$
ⓑ $\sqrt{s - 3} + 2 = \sqrt{s + 4}$

Show solution

ⓐ $no solution$ ⓑ $\frac{57}{16}$

ⓐ $\sqrt{u} + 1 = \sqrt{u + 4}$
ⓑ $\sqrt{n - 5} + 4 = \sqrt{3 n + 7}$

ⓐ $\sqrt{x} + 10 = \sqrt{x + 2}$
ⓑ $\sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Show solution

ⓐ $no solution$ ⓑ 6

$\sqrt{2 y + 4} + 6 = 0$

$\sqrt{8 u + 1} + 9 = 0$

Show solution

no solution

$\sqrt{a} + 1 = \sqrt{a + 5}$

$\sqrt{d} - 2 = \sqrt{d - 20}$

Show solution

$\sqrt{6 s + 4} = \sqrt{8 s - 28}$

$\sqrt{9 p + 9} = \sqrt{10 p - 6}$

Show solution

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To isolate the radical, add 9 to both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is n = 17.


To isolate the radical, subtract 2 from both sides.
Simplify.
Square both sides of the equation.
Solve the new equation.


Check the answer.

	The solution is $y = \frac{4}{3}$ .


To isolate the radical, subtract 1 from both sides.
Simplify.
Since the square root is equal to a negative number, the equation has no solution.