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Solve Equations with Square Roots

Solve radical equations

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 \(\sqrt{a}\ge 0\) 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.

Example

Try it.

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.

\[\text{For}\ a\ge 0,\ {(\sqrt{a})}^{2}=a\]

How to Solve Radical Equations

Try it.

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

Solution

Example

Try it.

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

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.

Example

Try it.

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

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 \(y=\frac{4}{3}\).

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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 \({s}^{2}\).

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.

\(\begin{array}{llll} & & & \ A={s}^{2} \\ \text{Take the square root of both sides.} & & & \sqrt{A}=\sqrt{{s}^{2}} \\ \text{Simplify.} & & & \sqrt{A}=s\end{array}\)

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.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

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 \(s=\sqrt{24d}\).

Solve Equations with Square Roots

Solve Radical Equations

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

Try it.

For the equation \(\sqrt{x+12}=x\): ⓐ Is \(x=4\) a solution? ⓑ Is \(x=-3\) a solution?

Solution

ⓐ \(\text{yes}\) ⓑ \(\text{no}\)

Try it.

For the equation \(\sqrt{\text{-}y+20}=y\): ⓐ Is \(y=4\) a solution? ⓑ Is \(y=-5\) a solution?

Try it.

For the equation \(\sqrt{t+6}=t\): ⓐ Is \(t=-2\) a solution? ⓑ Is \(t=3\) a solution?

Solution

ⓐ no ⓑ yes

Try it.

For the equation \(\sqrt{u+42}=u\): ⓐ Is \(u=-6\) a solution? ⓑ Is \(u=7\) a solution?

In the following exercises, solve.

Try it.

\(\sqrt{5y+1}=4\)

Solution

Try it.

\(\sqrt{7z+15}=6\)

Try it.

\(\sqrt{5x-6}=8\)

Solution

Try it.

\(\sqrt{4x-3}=7\)

Try it.

\(\sqrt{2m-3}-5=0\)

Solution

Try it.

\(\sqrt{2n-1}-3=0\)

Try it.

\(\sqrt{6v-2}-10=0\)

Solution

\(17\)

Try it.

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

Try it.

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

Solution

\(\frac{13}{5}\)

Try it.

\(\sqrt{4m+2}+2=6\)

Try it.

\(\sqrt{6n+1}+4=8\)

Solution

\(\frac{5}{2}\)

Try it.

\(\sqrt{2u-3}+2=0\)

Try it.

\(\sqrt{5v-2}+5=0\)

Solution

\(\text{no solution}\)

Try it.

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

Try it.

\(\sqrt{2m+1}+4=0\)

Solution

\(\text{no solution}\)

Try it.

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

Try it.

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

Solution

ⓐ \(10,11\) ⓑ \(5\)

Try it.

ⓐ \(\sqrt{r-1}-r=-1\)
ⓑ \(\sqrt{z+100}-z+10=0\)

Try it.

ⓐ \(\sqrt{s-8}-s=-8\)
ⓑ \(\sqrt{w+25}-w+5=0\)

Solution

ⓐ \(8,9\) ⓑ \(11\)

Try it.

\(3\sqrt{2x-3}-20=7\)

Try it.

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

Solution

\(3\)

Try it.

\(2\sqrt{8r+1}-8=2\)

Try it.

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

Solution

\(5\)

Try it.

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

Try it.

\(\sqrt{4v+3}=\sqrt{v-6}\)

Solution

not a real number

Try it.

\(\sqrt{8+2r}=\sqrt{3r+10}\)

Try it.

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

Solution

\(\frac{1}{4}\)

Try it.

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

Try it.

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

Solution

ⓐ \(\text{no solution}\) ⓑ \(\frac{57}{16}\)

Try it.

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

Try it.

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

Solution

ⓐ \(\text{no solution}\) ⓑ 6

Try it.

\(\sqrt{2y+4}+6=0\)

Try it.

\(\sqrt{8u+1}+9=0\)

Solution

no solution

Try it.

\(\sqrt{a}+1=\sqrt{a+5}\)

Try it.

\(\sqrt{d}-2=\sqrt{d-20}\)

Solution

Try it.

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

Try it.

\(\sqrt{9p+9}=\sqrt{10p-6}\)

Solution

Use Square Roots in Applications

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

Try it.

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.

Try it.

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

\(11.4\ \text{feet}\)

Try it.

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.

Try it.

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

\(8\ \text{seconds}\)

Try it.

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.

Try it.

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

\(15.8\ \text{seconds}\)

Try it.

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

Try it.

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

Solution

\(72\ \text{miles per hour}\)

Try it.

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 \(s=\sqrt{24d}\) to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Try it.

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 \(s=\sqrt{24d}\) to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest tenth.

Solution

\(53.0\ \text{miles per hour}\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: ⓐ \(\sqrt{9}\) ⓑ \({9}^{2}\).
If you missed this problem, review and .

Bonisa impendulo

ⓐ \(3\) ⓑ \(81\)
Solve: \(5(x+1)-4=3(2x-7)\).
If you missed this problem, review .

Bonisa impendulo

\(22\)
Solve: \({n}^{2}-6n+8=0\).
If you missed this problem, review .

Bonisa impendulo

\(n=2\ \text{or}\ n=4\)
For the equation \(\sqrt{x+2}=x\):
ⓐ Is \(x=2\) a solution? ⓑ Is \(x=-1\) a solution?

Bonisa impendulo

ⓐ 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.
For the equation \(\sqrt{x+6}=x\):
ⓐ Is \(x=-2\) a solution? ⓑ Is \(x=3\) a solution?

Bonisa impendulo

ⓐ no ⓑ \(\text{yes}\)
For the equation \(\sqrt{\text{-}x+2}=x\):
ⓐ Is \(x=-2\) a solution? ⓑ Is \(x=1\) a solution?

Bonisa impendulo

ⓐ no ⓑ \(\text{yes}\)
Solve: \(\sqrt{2x-1}=7\).
Solve: \(\sqrt{3x-5}=5\).

Bonisa impendulo

\(10\)
Solve: \(\sqrt{4x+8}=6\).

Bonisa impendulo

\(7\)
Solve: \(\sqrt{5n-4}-9=0\).

Bonisa impendulo

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: \(\sqrt{3m+2}-5=0\).

Bonisa impendulo

\(\frac{23}{3}\)
Solve: \(\sqrt{10z+1}-2=0\).

Bonisa impendulo

\(\frac{3}{10}\)
Solve: \(\sqrt{3y+5}+2=5\).

Bonisa impendulo

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 \(y=\frac{4}{3}\).
Solve: \(\sqrt{3p+3}+3=5\).

Bonisa impendulo

\(\frac{1}{3}\)
Solve: \(\sqrt{5q+1}+4=6\).

Bonisa impendulo

\(\frac{3}{5}\)
Solve: \(\sqrt{9k-2}+1=0\).

Bonisa impendulo

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.
Solve: \(\sqrt{2r-3}+5=0\).

Bonisa impendulo

\(\text{no solution}\)
Solve: \(\sqrt{7s-3}+2=0\).

Bonisa impendulo

\(\text{no solution}\)

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

Bonisa impendulo


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: \(\sqrt{x-2}+2=x\).

Bonisa impendulo

\(2,3\)
Solve: \(\sqrt{y-5}+5=y\).

Bonisa impendulo

\(5,6\)

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

Bonisa impendulo

	\(\sqrt{r+4}-r+2=0\)
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.	\(\ r+4={r}^{2}-4r+4\)
It is a quadratic equation, so get zero on one side.	\(\ 0={r}^{2}-5r\)
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.

Solve: \(\sqrt{m+9}-m+3=0\).

Bonisa impendulo

\(7\)
Solve: \(\sqrt{n+1}-n+1=0\).

Bonisa impendulo

\(3\)

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

Bonisa impendulo

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

Solve: \(2\sqrt{4a+2}-16=16\).

Bonisa impendulo

\(\frac{127}{2}\)
Solve: \(3\sqrt{6b+3}-25=50\).

Bonisa impendulo

\(\frac{311}{3}\)

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

Bonisa impendulo

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

Solve: \(\sqrt{2x-5}=\sqrt{5x+3}\).

Bonisa impendulo

no solution
Solve: \(\sqrt{7y+1}=\sqrt{2y-5}\).

Bonisa impendulo

no solution

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

Bonisa impendulo

	\(\ \sqrt{m}+1=\sqrt{m+9}\)
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.	\(\ 4m=64\)
	\(\ m=16\)
	Check the answer.
We leave it to you to show that \(m=16\) checks!	The solution is \(m=16.\)

Solve: \(\sqrt{x}+3=\sqrt{x+5}\).

Bonisa impendulo

\(\text{no solution}\)
Solve: \(\sqrt{m}+5=\sqrt{m+16}\).

Bonisa impendulo

\(\text{no solution}\)

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

Bonisa impendulo

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

Solve: \(\sqrt{y-3}+2=\sqrt{4y+2}\).

Bonisa impendulo

\(\text{no solution}\)
Solve: \(\sqrt{n-4}+5=\sqrt{3n+3}\).

Bonisa impendulo

\(\text{no solution}\)

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.

Bonisa impendulo

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.

Katie wants to plant a square lawn in her front yard. She has enough sod to cover an area of 370 square feet. Use the formula \(s=\sqrt{A}\) to find the length of each side of her lawn. Round your answer to the nearest tenth of a foot.

Bonisa impendulo

\(19.2\ \text{feet}\)
Sergio wants to make a square mosaic as an inlay for a table he is building. He has enough tile to cover an area of 2704 square centimeters. Use the formula \(s=\sqrt{A}\) to find the length of each side of his mosaic. Round your answer to the nearest tenth of a centimeter.

Bonisa impendulo

\(52.0\ \text{cm}\)

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.

Bonisa impendulo

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.
\(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.

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Equations with Square Roots

Solve radical equations
Use square roots in applications
Isolate the radical on one side of the equation.
Square both sides of the equation.
Solve the new equation.
Check the answer.
Isolate the radical on one side of the equation.
Square both sides of the equation.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama wena

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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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.