MX Algebra Solve Radical Equations

Section 8.6Solve Radical Equations

Definition

Before you get started, take this readiness quiz.

1

Simplify: \({(y-3)}^{2}.\)
If you missed this problem, review Example 7.

\({y}^{2}-6y+9\)

Definition
2

Solve: \(2x-5=0.\)
If you missed this problem, review Example 2.

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

Definition
3

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

\(n=2\,\text{or}\,n=4\)

Solve Radical Equations

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation.

Radical Equation

An equation in which a variable is in the radicand of a radical expression is called a radical equation.

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the nth power. This will eliminate the radical.

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

Example 1How to Solve a Radical Equation

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

Isolate the radical on one side of the equation before squaring both sides.

Step 1 is to isolate the radical on one side of the equation. To isolate the radical add 9 to both sides. The resulting equation is square root of the quantity 5 n minus 4 in parentheses minus 9 plus 9 equals 0 plus 9. This simplifies to square root of the quantity 5 n minus 4 in parentheses equals 9. Step 2 is to raise both sides of the equation to the power of the index. Since the index of a square root is 2, we square both sides. Remember that the square of the square root of “a” is equal to “a”. The equation that results is the square of the square root of the quantity 5 n minus 4 in parentheses equals 9 squared. This simplifies to 5 n minus 4 equals 81. Step 3 is to solve the new equation. We get 5 n equals 85 and then n equals 17. Step 4 is to check the answer in the original equation. Does the square root of the quantity 5 times 17 minus 4 in parentheses minus 9 equal zero? Simplifying the left side we get square root of the quantity 85 minus 4 in parentheses minus 9 and then square root of 81 minus 9 and then 9 minus 9 which does equal 0. This verifies that the solution is n equals 17.
Try It #1

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

\(m=\frac{23}{3}\)

Did you get it?
Try It #2

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

\(z=\frac{3}{10}\)

Did you get it?
Solve a radical equation with one radical.
  • Isolate the radical on one side of the equation.
  • Raise both sides of the equation to the power of the index.
  • Solve the new equation.
  • Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Example 2

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

Isolate the radical first, then notice what value it would have to equal.

Table 1
A mathematical equation is displayed, showing the square root of (9k - 2), with a plus 1 outside the square root, set equal to 0. The full equation is: sqrt(9k - 2) + 1 = 0.
To isolate the radical, subtract 1 to both sides.A mathematical equation: the square root of 9k minus 2, plus 1, minus 1, equals 0 minus 1. The '-1' terms are highlighted in red on both sides of the equation.
Simplify.A mathematical equation displays the square root of 9k minus 2, which equals negative 1. The equation is 'sqrt(9k - 2) = -1'.

Because the square root is equal to a negative number, the equation has no solution.

Try It #3

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

\(\text{no solution}\)

Did you get it?
Try It #4

Solve: \(\sqrt{7s-3}+2=0.\)

\(\text{no solution}\)

Did you get it?

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Binomial Squares

\[\begin{array}{l} \\ \\ {(a+b)}^{2}={a}^{2}+2ab+{b}^{2} \\ {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}\]

Don’t forget the middle term!

Example 3

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

Isolate the radical, then square both sides, using the binomial squares pattern on the right side.

Table 2
A mathematical equation is displayed, showing 'sqrt(p - 1) + 1 = p'.
To isolate the radical, subtract 1 from both sides.An equation, sqrt(p-1) + 1 - 1 = p-1, featuring mathematical symbols and numbers, with the second '-1' and the 'p-1' on the right side highlighted in red.
Simplify.A mathematical equation is displayed, showing the square root of (p-1) equal to (p-1).
Square both sides of the equation.The equation shows (the square root of p-1) squared equals (p-1) squared.
Simplify, using the Product of Binomial Squares Pattern on the
right. Then solve the new equation.		A mathematical equation is displayed on a white background, which reads 'p - 1 = p^2 - 2p + 1'.
It is a quadratic equation, so get zero on one side.A quadratic equation showing zero equals the quantity p squared minus three p plus two.
Factor the right side.A mathematical equation showing 0 equals the product of (p - 1) and (p - 2).
Use the Zero Product Property.Two mathematical equations are displayed horizontally: '0 = p - 1' and '0 = p - 2'. The background is white.
Solve each equation.Two mathematical expressions are displayed on a white background: 'p = 1' on the left and 'p = 2' on the right, indicating two distinct parameter values.
Check the answers.
Step-by-step verification of the equation sqrt(p-1) + 1 = p, confirming its truth for p=1 and p=2 in these two examples.
The solutions are \(p=1,\,p=2.\)
Try It #5

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

\(x=2,x=3\)

Did you get it?
Try It #6

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

\(y=5,y=6\)

Did you get it?

When the index of the radical is 3, we cube both sides to remove the radical.

\[{(\sqrt[3]{a})}^{3}=a\]

Example 4

Solve: \(\sqrt[3]{5x+1}+8=4.\)

Isolate the radical, then cube both sides since the index is 3.

Table 3
\(\sqrt[3]{5x+1}+8=4\,\)
To isolate the radical, subtract 8 from both sides.\(\sqrt[3]{5x+1}=-4\,\)
Cube both sides of the equation.\({(\sqrt[3]{5x+1})}^{3}={(-4)}^{3}\)
Simplify.\(5x+1=-64\,\)
Solve the equation.\(5x=-65\,\)
\(x=-13\,\)
Check the answer.
Verifying that x = -13 is the correct solution for the equation cube_root(5x + 1) + 8 = 4, demonstrating step-by-step substitution and simplification to confirm the equality 4 = 4.
The solution is \(x=-13.\,\)
Try It #7

Solve: \(\sqrt[3]{4x-3}+8=5\)

\(x=-6\)

Did you get it?
Try It #8

Solve: \(\sqrt[3]{6x-10}+1=-3\)

\(x=-9\)

Did you get it?

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since \({({a}^{m})}^{n}={a}^{m·n},\) we have for example,

\[{({x}^{\frac{1}{2}})}^{2}=x,\,{({x}^{\frac{1}{3}})}^{3}=x\]

Remember, \({x}^{\frac{1}{2}}=\sqrt{x}\) and \({x}^{\frac{1}{3}}=\sqrt[3]{x}.\)

Example 5

Solve: \({(3x-2)}^{\frac{1}{4}}+3=5.\)

Isolate the term with the rational exponent, then raise both sides to the fourth power.

Table 4
\({(3x-2)}^{\frac{1}{4}}+3=5\,\)
To isolate the term with the rational exponent,
subtract 3 from both sides.		\({(3x-2)}^{\frac{1}{4}}=2\,\)
Raise each side of the equation to the fourth power.\({({(3x-2)}^{\frac{1}{4}})}^{4}={(2)}^{4}\)
Simplify.\(3x-2=16\,\)
Solve the equation.\(3x=18\,\)
\(x=6\,\)
Check the answer.
Verifying x=6 as the solution to (3x-2)^(1/4) + 3 = 5 by substituting x and simplifying the equation to confirm 5=5.
The solution is \(x=6.\,\)
Try It #9

Solve: \({(9x+9)}^{\frac{1}{4}}-2=1.\)

\(x=8\)

Did you get it?
Try It #10

Solve: \({(4x-8)}^{\frac{1}{4}}+5=7.\)

\(x=6\)

Did you get it?

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution!

Example 6

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

Isolate the radical and square both sides, then check every solution in the original equation for extraneous roots.

Table 5
\(\sqrt{r+4}-r+2=0\,\)
Isolate the radical.\(\sqrt{r+4}=r-2\,\)
Square both sides of the equation.\({(\sqrt{r+4})}^{2}={(r-2)}^{2}\,\)
Simplify and then solve the 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 your answer.
Verification of solutions for the radical equation sqrt(r+4) - r + 2 = 0. It shows that r=0 is an extraneous solution, while r=5 is a valid solution. The solution is r = 5.
\(r=0\) is an extraneous solution.
Try It #11

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

\(m=7\)

Did you get it?
Try It #12

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

\(n=3\)

Did you get it?

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Example 7

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

Isolate the radical term completely, including dividing out its coefficient, before squaring.

Table 6
\(\text{3}\,\sqrt{3x-5}-8=4\,\)
Isolate the radical term.\(3\sqrt{3x-5}=12\,\)
Isolate the radical by dividing both sides by 3.\(\sqrt{3x-5}=4\,\)
Square both sides of the equation.\({(\sqrt{3x-5})}^{2}={(4)}^{2}\)
Simplify, then solve the new equation.\(3x-5=16\,\)
\(3x=21\,\)
Solve the equation.\(x=7\,\)
Check the answer.
This image demonstrates the step-by-step verification that x=7 is a valid solution to the radical equation 3√(3x-5) - 8 = 4, successfully leading to 4=4.
The solution is \(x=7.\,\)
Try It #13

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

\(a=63\)

Did you get it?
Try It #14

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

\(b=311\)

Did you get it?

Solve Radical Equations with Two Radicals

If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.

In the next example, when one radical is isolated, the second radical is also isolated.

Example 8

Solve: \(\sqrt[3]{4x-3}=\sqrt[3]{3x+2}.\)

Both radicals are already isolated, so cube both sides right away to remove them.

Table 7
The radical terms are isolated.\(\,\sqrt[3]{4x-3}=\sqrt[3]{3x+2}\)
Since the index is 3, cube both sides of the
equation.		\(\,{(\sqrt[3]{4x-3})}^{3}={(\sqrt[3]{3x+2})}^{3}\)
Simplify, then solve the new equation.\(\,4x-3=3x+2\)
\(\,x-3=2\)
\(\,x=5\)
\(\text{The solution is}\,x=5.\)
Check the answer.
We leave it to you to show that 5 checks!
Try It #15

Solve: \(\sqrt[3]{5x-4}=\sqrt[3]{2x+5}.\)

\(x=3\)

Did you get it?
Try It #16

Solve: \(\sqrt[3]{7x+1}=\sqrt[3]{2x-5}.\)

\(x=-\frac{6}{5}\)

Did you get it?

Sometimes after raising both sides of an equation to a power, 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 raise both sides of the equation to the power of the index again.

Example 9How to Solve a Radical Equation

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

One radical is already isolated — square both sides, then isolate and square the radical that remains.

Step 1 is to isolate one of the radical terms on one side of the equation. The radical on the right is isolated. Step 2 is to raise both sides of the equation to the power of the index. We square both sides. The equation that results is the square of the quantity square root of m plus 1 in parentheses equals the square of the square root of the quantity m plus 9 in parentheses. Simplify – be very careful as you multiply! This simplifies to m plus 2 times square root m plus 1 equals m plus 9. Step 3 is to repeat steps 1 and 2 again if there are any more radicals. There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical term. 2 times square root m equals 8. Here, we can easily isolate the radical by dividing both sides by 2. We get square root m equals 4. Squaring both sides we get the square of the square root of m equals 4 squared. m equals 16. Step 4 is to check the answer in the original equation. Does the square root of 16 plus 1 equal the square root of the quantity 16 plus 9? Simplifying both sides we get 4 plus 1 equals 5. This verifies that the solution is m equals 16.
Try It #17

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

\(x=4\)

Did you get it?
Try It #18

Solve: \(\sqrt{x}+2=\sqrt{x+16}.\)

\(x=9\)

Did you get it?

We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.

Solve a radical equation.
  • Isolate one of the radical terms on one side of the equation.
  • Raise both sides of the equation to the power of the index.
  • Are there any more radicals?
    If yes, repeat Step 1 and Step 2 again.
    If no, solve the new equation.
  • Check the answer in the original equation.

Be careful as you square binomials in the next example. Remember the pattern is \({(a+b)}^{2}={a}^{2}+2ab+{b}^{2}\) or \({(a-b)}^{2}={a}^{2}-2ab+{b}^{2}.\)

Example 10

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

Square both sides using the binomial squares pattern on the left, then isolate and square the radical that remains.

Table 8
A mathematical equation shows the square root of (q - 2) plus 3 equals the square root of (4q + 1).
The radical on the right is isolated. Square
both sides.		An algebraic equation showing the square of the expression (sqrt(q-2) + 3) on the left side, and the square of sqrt(4q+1) on the right side, with both expressions being equal.
Simplify.A mathematical equation is displayed: q - 2 + 6 * sqrt(q - 2 + 9) = 4q + 1. The equation involves a variable 'q', numbers, addition, subtraction, multiplication, and a square root.
There is still a radical in the equation so
we must repeat the previous steps. Isolate
the radical.		A mathematical equation is displayed on a white background, which reads '6 sqrt(q) - 2 = 3q - 6'.
Square both sides. It would not help to
divide both sides by 6. Remember to
square both the 6 and the \(\sqrt{q-2}.\)		The image shows mathematical equations demonstrating algebraic expansions. The top line presents two squared expressions: (6√(q-2))^2 and ((a-b)/(3q-6))^2. The bottom line details the expansion of 6^2(√(q-2))^2 and the binomial expansion of (3q-6)^2, referencing the identity a^2 - 2ab + b^2.
Simplify, then solve the new equation.A mathematical equation is displayed, showing 36(q-2) = 9q^2 - 36q + 36 on a white background.
Distribute.A quadratic equation is displayed, showing 36q - 72 = 9q^2 - 36q + 36, likely for solving the variable 'q'.
It is a quadratic equation, so get zero on
one side.		A quadratic equation is displayed: 0 = 9q^2 - 72q + 108. The equation is presented in a clear, digital format against a white background.
Factor the right side.Algebraic factorization of a quadratic expression. The equation 0 = 9(q^2 - 8q + 12) is factored into 0 = 9(q - 6)(q - 2).
Use the Zero Product Property.Two separate sets of linear equations and their solutions are shown. On the left, q - 6 = 0 is solved to q = 6. On the right, q - 2 = 0 is solved to q = 2.
The checks are left to you.The solutions are \(q=6\) and \(q=2.\)
Try It #19

Solve: \(\sqrt{x-1}+2=\sqrt{2x+6}\)

\(x=5\)

Did you get it?
Try It #20

Solve: \(\sqrt{x}+2=\sqrt{3x+4}\)

\(x=0\,x=4\)

Did you get it?

Use Radicals in Applications

As you progress through your college courses, you’ll encounter formulas that include radicals in many disciplines. We will modify our Problem Solving Strategy for Geometry Applications slightly to give us a plan for solving applications with formulas from any discipline.

Use a problem solving strategy for applications with formulas.
  • 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.
  • Identify what we are looking for.
  • Name what we are looking for by choosing a variable to represent it.
  • Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  • Solve the equation using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

One application of radicals has to do with the effect of gravity on falling objects. The formula allows us to determine how long it will take a fallen object to hit the gound.

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

\[t=\frac{\sqrt{h}}{4}.\]

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.

Table 9
A mathematical formula displaying 't equals the square root of h, all divided by 4', which can also be written as t = vh / 4 or t = (h^0.5) / 4.
A mathematical equation: t equals the square root of 64, all divided by 4, with the number 64 highlighted in red.
Take the square root of 64.A mathematical equation shows 't = 8/4' in black text against a white background.
Simplify the fraction.The text 't=2' is displayed in dark gray characters against a plain white background.

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

Example 11

Marissa 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.

Substitute \(h=400\) into the formula and simplify the square root.

Table 10
Step 1. Read the problem.
Step 2. Identify what we are looking for.the time it takes for the
sunglasses to reach the river
Step 3. Name what we are looking.Let \(t=\) time.
Step 4. Translate into an equation by writing the
appropriate formula. Substitute in the given
information.		Substitution of h=400 into the equation t = sqrt(h)/4, resulting in t = sqrt(400)/4, demonstrating the first step in solving for t.
Step 5. Solve the equation.A mathematical expression reads 't = 20/4', indicating a variable 't' is equal to the fraction 20 divided by 4.
The image displays the mathematical expression 't = 5' in black text against a plain white background, indicating a variable 't' assigned the value of five.
Step 6. Check the answer in the problem and make
sure it makes sense.		A three-step math problem demonstrating that 5 is equal to the square root of 400 divided by 4, concluding with '5 = 5' and a checkmark.
Does 5 seconds seem like a reasonable length of
time?		Yes.
Step 7. Answer the question.It will take 5 seconds for the
sunglasses to reach the river.
Try It #21

A helicopter dropped a rescue package from a height of 1,296 feet. Use the formula \(t=\frac{\sqrt{h}}{4}\) to find how many seconds it took for the package to reach the ground.

9 seconds

Did you get it?
Try It #22

A window washer dropped a squeegee from a platform 196 feet above the sidewalk Use the formula \(t=\frac{\sqrt{h}}{4}\) to find how many seconds it took for the squeegee to reach the sidewalk.

\(3.5\) seconds

Did you get it?

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.

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

\[s=\sqrt{24d}\]

Example 12

After a car accident, the skid marks for one car measured 190 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.

Substitute \(d=190\) into the formula and evaluate the square root.

Table 11
Step 1. Read the problem
Step 2. Identify what we are looking for.the speed of a car
Step 3. Name what weare looking for,Let \(s=\) the speed.
Step 4. Translate into an equation by writing
the appropriate formula. Substitute in the
given information.		A mathematical problem showing the substitution of d=190 into the equation s = sqrt(24d), resulting in s = sqrt(24(190)), with 190 highlighted in red.
Step 5. Solve the equation.The image displays the mathematical equation s = sqrt(4,560), calculating the square root of 4,560.
The variable 's' is equal to the repeating decimal 67.52777...
Round to 1 decimal place.A mathematical expression displays 's' is approximately equal to 67.5.
Checking if 67.5 equals the square root of 24 times 190. Steps show simplification: square root of 24 times 190 becomes square root of 4560, which is approximately 67.5277. A checkmark confirms the approximation.
The speed of the car before the brakes were applied
was 67.5 miles per hour.
Try It #23

An accident investigator measured the skid marks of the car. The length of the skid marks was 76 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.

\(42.7\) feet

Did you get it?
Try It #24

The skid marks of a vehicle involved in an accident were 122 feet long. 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.

\(54.1\) feet

Did you get it?

Key Concepts

Section Exercises

Practice Makes Perfect

Solve Radical Equations

In the following exercises, solve.

4

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

\(x=14\)

5

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

6

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

no solution

7

\(\sqrt{3y-4}=-2\)

8

\(\sqrt[3]{2x}=-2\)

\(x=-4\)

9

\(\sqrt[3]{4x-1}=3\)

10

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

\(m=14\)

11

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

12

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

\(v=17\)

13

\(\sqrt{12u+1}-11=0\)

14

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

\(m=\frac{7}{2}\)

15

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

16

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

no solution

17

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

18

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

\(u=3,u=4\)

19

\(\sqrt{v-10}+10=v\)

20

\(\sqrt{r-1}=r-1\)

\(r=1,r=2\)

21

\(\sqrt{s-8}=s-8\)

22

\(\sqrt[3]{6x+4}=4\)

\(x=10\)

23

\(\sqrt[3]{11x+4}=5\)

24

\(\sqrt[3]{4x+5}-2=-5\)

\(x=-8\)

25

\(\sqrt[3]{9x-1}-1=-5\)

26

\({(6x+1)}^{\frac{1}{2}}-3=4\)

\(x=8\)

27

\({(3x-2)}^{\frac{1}{2}}+1=6\)

28

\({(8x+5)}^{\frac{1}{3}}+2=-1\)

\(x=-4\)

29

\({(12x-5)}^{\frac{1}{3}}+8=3\)

30

\({(12x-3)}^{\frac{1}{4}}-5=-2\)

\(x=7\)

31

\({(5x-4)}^{\frac{1}{4}}+7=9\)

32

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

\(x=3\)

33

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

34

\(\sqrt{z+100}-z=-10\)

\(z=21\)

35

\(\sqrt{w+25}-w=-5\)

36

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

\(x=42\)

37

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

38

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

\(r=3\)

39

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

Solve Radical Equations with Two Radicals

In the following exercises, solve.

40

\(\sqrt{3u+7}=\sqrt{5u+1}\)

\(u=3\)

41

\(\sqrt{4v+1}=\sqrt{3v+3}\)

42

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

\(r=-2\)

43

\(\sqrt{10+2c}=\sqrt{4c+16}\)

44

\(\sqrt[3]{5x-1}=\sqrt[3]{x+3}\)

\(x=1\)

45

\(\sqrt[3]{8x-5}=\sqrt[3]{3x+5}\)

46

\(\sqrt[3]{2{x}^{2}+9x-18}=\sqrt[3]{{x}^{2}+3x-2}\)

\(x=-8,x=2\)

47

\(\sqrt[3]{{x}^{2}-x+18}=\sqrt[3]{2{x}^{2}-3x-6}\)

48

\(\sqrt{a}+2=\sqrt{a+4}\)

\(a=0\)

49

\(\sqrt{r}+6=\sqrt{r+8}\)

50

\(\sqrt{u}+1=\sqrt{u+4}\)

\(u=\frac{9}{4}\)

51

\(\sqrt{x}+1=\sqrt{x+2}\)

52

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

\(a=4\)

53

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

54

\(\sqrt{2x+1}=1+\sqrt{x}\)

\(x=0\,x=4\)

55

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

56

\(\sqrt{2x-1}-\sqrt{x-1}=1\)

\(x=1\,x=5\)

57

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

58

\(\sqrt{x+7}-\sqrt{x-5}=2\)

\(x=9\)

59

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

Use Radicals in Applications

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

60

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.

\(8.7\) feet

61

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.

62

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.

\(4.7\) seconds

63

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.

64

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.

72 feet

65

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.

Writing Exercises

66

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

Answers will vary.

67

ⓐ 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.

Self Check

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

The table has 4 columns and 4 rows. The first row is a header row with the headers “I can…”, “Confidently”, “With some help.”, and “No – I don’t get it!”. The first column contains the phrases “Solve radical equations”, “solve radical equations with two radicals”, and “use radicals in applications”. The other columns are left blank so the learner can indicate their level of understanding.

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

Glossary

radical equation
An equation in which a variable is in the radicand of a radical expression is called a radical equation.