8.6 Solve Radical Equations

Learning objectives

Learning objectives

• Solve radical equations

• Solve radical equations with two radicals

• Use radicals in applications

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.

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 n${}^{th}$ power. This will eliminate the radical.

Example 8.56

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 8.57

Solution

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

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.

Don’t forget the middle term!

Example 8.58

Solution

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

Example 8.59

Solution

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 ${({a}^{m})}^{n}={a}^{m\cdot{}n},$ we have for example,

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

Example 8.60

Solution

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

Example 8.61

Solution

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

Example 8.62

Solution

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 8.63

Solution

The radical terms are isolated. · $\,\sqrt[3]{4x-3}=\sqrt[3]{3x+2}$ Since the index is 3, cube both sides of theequation. · $\,{(\sqrt[3]{4x-3})}^{3}={(\sqrt[3]{3x+2})}^{3}$ Simplify, then solve the new equation. · $\,4x-3=3x+2$

$\,x-3=2$

$\,x=5$

$\mathrm{The solution is}\,x=5.$

Check the answer.

We leave it to you to show that 5 checks!

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 8.64

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.

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 8.65

Solution

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.

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.

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.

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

Example 8.66

Solution

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.

Example 8.67

Solution

Section 8.6 Exercises

Section 8.6 Exercises

Solve Radical Equations

In the following exercises, solve.

Solve Radical Equations with Two Radicals

In the following exercises, solve.

Use Radicals in Applications

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

Writing Exercises

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

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

1. Solve: $\sqrt{3m+2}-5=0.$

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

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

4. Solve: $\sqrt{7s-3}+2=0.$

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

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

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

4. Solve: $\sqrt[3]{6x-10}+1=-3$

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

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

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

4. Solve: $\sqrt{n+1}-n+1=0.$

1. Solve: $2\sqrt{4a+4}-16=16.$

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

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

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

1. Solve: $3-\sqrt{x}=\sqrt{x-3}.$

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

3. Solve: $\sqrt{x-1}+2=\sqrt{2x+6}$

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

1. 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.

2. 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. 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.

4. 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.

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

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

3. $\sqrt[3]{2x}=-2$

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

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

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

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

4. $\sqrt{u-3}+3=u$

1. $\sqrt{r-1}=r-1$

2. $\sqrt[3]{6x+4}=4$

3. $\sqrt[3]{4x+5}-2=-5$

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

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

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

3. $\sqrt{x+1}-x+1=0$

4. $\sqrt{z+100}-z=-10$

1. $3\sqrt{2x-3}-20=7$

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

3. $\sqrt{3u+7}=\sqrt{5u+1}$

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

1. $\sqrt[3]{5x-1}=\sqrt[3]{x+3}$

2. $\sqrt[3]{2{x}^{2}+9x-18}=\sqrt[3]{{x}^{2}+3x-2}$

3. $\sqrt{a}+2=\sqrt{a+4}$

4. $\sqrt{u}+1=\sqrt{u+4}$

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

2. $\sqrt{2x+1}=1+\sqrt{x}$

3. $\sqrt{2x-1}-\sqrt{x-1}=1$

4. $\sqrt{x+7}-\sqrt{x-5}=2$

1. 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.

2. 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.

3. 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.

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