Solve Equations with Square Roots

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 $\sqrt{a} \geq 0$ 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. We saw extraneous solutions when we solved rational equations, too.

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

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

Solution

ⓐ Is

x = 2

a solution?


Let x = 2.
Simplify.

	2 is a solution.

ⓑ Is

x = −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

How to Solve Radical Equations

Solve: $\sqrt{2 x - 1} = 7$ .

Solution

This table has three columns and four rows. The first row says, “Step 1. Isolate the radical on one side of equation. The square root of (2x minus 1) is already isolated on the left side.” It then shows the equation: the square root of (2x minus 1) equals 7.

The second row says, “Step 2. Square both sides of the equation. Remember, the square root of a squared equals a.” It then shows the equation: the square root of (2x minus 1) squared equals 7 squared.

The third row then says, “Step 3. Solve the new equation.” It indicates that 2x minus 1 equals 49 or 2x equals 50 which means that x equals 25.

The fourth row says, “Step 4. Check the answer. Check:” It then indicates the square root of (2x minus 1) equals 7. This becomes the square root of (2 times 25 minus 1) equals 7. This becomes the square root of (50 minus 1) equals 7. This becomes the square root of 49 equals 7, and thus 7 equals 7. The figure then states, “The solutions is x equals 25.”

Solve: $\sqrt{5 n - 4} - 9 = 0$ .

Solution


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.

Solve: $\sqrt{3 y + 5} + 2 = 5$ .

Solution


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 $y = \frac{4}{3}$ .

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

Solve: $\sqrt{9 k - 2} + 1 = 0$ .

Solution


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.

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!

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

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.

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

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 $r = 5$ .
	$r = 0$ is an extraneous solution.

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

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

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.	$27 x = 189$
	$x = 7$
Check the answer.
	The solution is $x = 7$ .

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

Solution

Step-by-step solution demonstrating how to solve a radical equation by isolating and squaring radical terms.
	$\begin{matrix} \sqrt{4 z - 3} = \sqrt{3 z + 2} \end{matrix}$
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 $z = 5 .$

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.

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

Solution

Step-by-step solution for the radical equation "sqrt(m) + 1 = sqrt(m + 9)", illustrating the process of isolating and squaring radicals to find the value of 'm'.
	$\sqrt{m} + 1 = \sqrt{m + 9}$
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.	$4 m = 64$
	$m = 16$
	Check the answer.
We leave it to you to show that $m = 16$ checks!	The solution is $m = 16 .$

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

Solution

Step-by-step solution demonstrating how to solve a radical equation by isolating and squaring radical terms to find the values of q.
	$\sqrt{q - 2} + 3 = \sqrt{4 q + 1}$
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.)	$The solutions are q = 6 and q = 2 .$

Use Square Roots in Applications

As you progress through your college courses, you’ll encounter formulas that include square roots in many disciplines. We have already used formulas to solve geometry applications.

We will use our Problem Solving Strategy for Geometry Applications, with slight modifications, to give us a plan for solving applications with formulas from any discipline.

We used the formula $A = L \cdot W$ to find the area of a rectangle with length L and width W. A square is a rectangle in which the length and width are equal. If we let s be the length of a side of a square, the area of the square is $s^{2}$ .

This figure shows a square with two sides labeled s. It also indicates that A equals s squared.

The formula $A = s^{2}$ gives us the area of a square if we know the length of a side. What if we want to find the length of a side for a given area? Then we need to solve the equation for s.

$\begin{matrix} A = s^{2} \\ Take the square root of both sides. & \sqrt{A} = \sqrt{s^{2}} \\ Simplify. & \sqrt{A} = s \end{matrix}$

We can use the formula $s = \sqrt{A}$ to find the length of a side of a square for a given area.

We will show an example of this in the next example.

Mike and Lychelle want to make a square patio. They have enough concrete to pave an area of 200 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of the patio. Round your answer to the nearest tenth of a foot.

Solution

Step 1. Read the problem. Draw a figure and label it with the given information.
	A = 200 square feet
Step 2. Identify what you are looking for.	The length of a side of the square patio.
Step 3. Name what you are looking for by choosing a variable to represent it.	Let s = the length of a side.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute the given information.
Step 5. Solve the equation using good algebra techniques. Round to one decimal place.
Step 6. Check the answer in the problem and make sure it makes sense.

This is close enough because we rounded the square root. Is a patio with side 14.1 feet reasonable? Yes.
Step 7. Answer the question with a complete sentence.	Each side of the patio should be 14.1 feet.

Another application of square roots has to do with gravity.

For example, if an object is dropped from a height of 64 feet, we can find the time it takes to reach the ground by substituting $h = 64$ into the formula.



Take the square root of 64.	$A mathematical equation displays 't = 8/4' on a white background, representing the variable t being equal to the fraction eight divided by four.$
Simplify the fraction.

It would take 2 seconds for an object dropped from a height of 64 feet to reach the ground.

Christy dropped her sunglasses from a bridge 400 feet above a river. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the sunglasses to reach the river.

Solution

Step 1. Read the problem.
Step 2. Identify what you are looking for.	The time it takes for the sunglasses to reach the river.
Step 3. Name what you are looking for by choosing a variable to represent it.	Let t = time.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
$A math equation displays 5 followed by a question mark over an equals sign, then the fraction 20 over 4, asking if 5 is equal to 20 divided by 4.$ $5 = 5 ✓$
Does 5 seconds seem reasonable? Yes.
Step 7. Answer the question with a complete sentence.	It will take 5 seconds for the sunglasses to hit the water.

Police officers investigating car accidents measure the length of the skid marks on the pavement. Then they use square roots to determine the speed, in miles per hour, a car was going before applying the brakes.

After a car accident, the skid marks for one car measured 190 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

Step 1. Read the problem.
Step 2. Identify what we are looking for.	The speed of a car.
Step 3. Name what we are looking for.	Let s = the speed.
Step 4. Translate into an equation by writing the appropriate formula.
Substitute the given information.
Step 5. Solve the equation.

Round to 1 decimal place.
Step 6. Check the answer in the problem. $67.5 \overset{?}{\approx} \sqrt{24 (190)}$ $67.5 \overset{?}{\approx} \sqrt{4560}$ $67.5 \overset{?}{\approx} 67.5277...$
Is 67.5 mph a reasonable speed?	Yes.
Step 7. Answer the question with a complete sentence.	The speed of the car was approximately 67.5 miles per hour.

Key Concepts

To Solve a Radical Equation:
1. Isolate the radical on one side of the equation.
2. Square both sides of the equation.
3. Solve the new equation.
4. Check the answer. Some solutions obtained may not work in the original equation.
Solving Applications with Formulas
1. Read the problem and make sure all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
2. Identify what we are looking for.
3. Name what we are looking for by choosing a variable to represent it.
4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
Area of a Square
Falling Objects
- On Earth, if an object is dropped from a height of $h$ feet, the time in seconds it will take to reach the ground is found by using the formula $t = \frac{\sqrt{h}}{4}$ .
Skid Marks and Speed of a Car
- If the length of the skid marks is d feet, then the speed, s, of the car before the brakes were applied can be found by using the formula $s = \sqrt{24 d}$ .

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 $x = 4$ a solution? ⓑ Is $x = −3$ a solution?

Solution

ⓐ $yes$ ⓑ $no$

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

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

Solution

ⓐ no ⓑ yes

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

In the following exercises, solve.

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

Solution

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

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

Solution

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

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

Solution

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

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

Solution

$17$

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

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

Solution

$\frac{13}{5}$

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

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

Solution

$\frac{5}{2}$

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

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

Solution

$no solution$

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

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

Solution

$no solution$

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

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

Solution

ⓐ $10, 11$ ⓑ $5$

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

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

Solution

ⓐ $8, 9$ ⓑ $11$

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

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

Solution

$3$

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

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

Solution

$5$

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

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

Solution

not a real number

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

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

Solution

$\frac{1}{4}$

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

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

Solution

ⓐ $no solution$ ⓑ $\frac{57}{16}$

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

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

Solution

ⓐ $no solution$ ⓑ 6

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

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

Solution

no solution

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

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

Solution

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

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

Solution

Use Square Roots in Applications

In the following exercises, solve. Round approximations to one decimal place.

Landscaping Reed wants to have a square garden plot in his backyard. He has enough compost to cover an area of 75 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his garden. Round your answer to the nearest tenth of a foot.

Landscaping Vince wants to make a square patio in his yard. He has enough concrete to pave an area of 130 square feet. Use the formula $s = \sqrt{A}$ to find the length of each side of his patio. Round your answer to the nearest tenth of a foot.

Solution

$11.4 feet$

Gravity While putting up holiday decorations, Renee dropped a light bulb from the top of a 64 foot tall tree. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the light bulb to reach the ground.

Gravity An airplane dropped a flare from a height of 1024 feet above a lake. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the flare to reach the water.

Solution

$8 seconds$

Gravity A hang glider dropped his cell phone from a height of 350 feet. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the cell phone to reach the ground.

Gravity A construction worker dropped a hammer while building the Grand Canyon skywalk, 4000 feet above the Colorado River. Use the formula $t = \frac{\sqrt{h}}{4}$ to find how many seconds it took for the hammer to reach the river.

Solution

$15.8 seconds$

Accident investigation The skid marks for a car involved in an accident measured 54 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation The skid marks for a car involved in an accident measured 216 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the car before the brakes were applied. Round your answer to the nearest tenth.

Solution

$72 miles per hour$

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 175 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Accident investigation An accident investigator measured the skid marks of one of the vehicles involved in an accident. The length of the skid marks was 117 feet. Use the formula $s = \sqrt{24 d}$ to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

$53.0 miles per hour$

Writing Exercises

Explain why an equation of the form $\sqrt{x} + 1 = 0$ has no solution.

ⓐ Solve the equation $\sqrt{r + 4} - r + 2 = 0$ .
ⓑ Explain why one of the “solutions” that was found was not actually a solution to the equation.

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has two rows and four columns. The first row labels each column, “I can…,” “Confidently,” “With some help,” and “No minus I don’t get it!” The row under “I can…,” reads, “use square roots in applications.” All the other rows are empty.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

radical equation

An equation in which the variable is in the radicand of a square root is called a radical equation