What Is A Radical Equation

Roots and Radicals

Solve Radical Equations

Learning Objectives

Past the end of this section, you lot will be able to:

Solve radical equations
Solve radical equations with ii radicals
Utilise radicals in applications

Before you get started, take this readiness quiz.

Simplify: ${\left(y-3\right)}^{2}.$
If you lot missed this problem, review (Figure).
Solve: $2x-5=0.$
If you lot missed this problem, review (Effigy).
Solve ${n}^{2}-6n+8=0.$
If you missed this problem, review (Figure).

Solve Radical Equations

In this section we volition solve equations that accept 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 chosen a radical equation.

Every bit usual, when solving these equations, what we do to one side of an equation we must practice to the other side likewise. 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. Once more, nosotros call this an extraneous solution as nosotros did when we solved rational equations.

In the side by side example, we will encounter how to solve a radical equation. Our strategy is based on raising a radical with alphabetize northward to the due north ^th power. This volition eliminate the radical.

$\text{For}\phantom{\rule{0.2em}{0ex}}a\ge 0,\phantom{\rule{0.2em}{0ex}}{\left(\sqrt[n]{a}\right)}^{n}=a.$

How to Solve a Radical Equation

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

$m=\frac{23}{3}$

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

$z=\frac{3}{10}$

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.
Bank 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 volition take no solution.

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

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

$\text{no solution}$

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

$\text{no solution}$

If one side of an equation with a square root is a binomial, we use the Production of Binomial Squares Pattern when nosotros square information technology.

Binomial Squares

$\begin{array}{}\\ \\ {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}$

Don't forget the middle term!

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

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

$x=2,x=3$

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

$y=5,y=6$

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

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

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

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

$x=-6$

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

$x=-9$

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

${\left({x}^{\frac{1}{2}}\right)}^{2}=x,\phantom{\rule{0.5em}{0ex}}{\left({x}^{\frac{1}{3}}\right)}^{3}=x$

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

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

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

$x=8$

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

$x=6$

Sometimes the solution of a radical equation results in ii algebraic solutions, just i of them may be an extraneous solution!

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

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

$m=7$

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

$n=3$

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

Solve: $\text{3}\phantom{\rule{0.2em}{0ex}}\sqrt{3x-5}-8=4.$

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

$a=63$

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

$b=311$

Solve Radical Equations with 2 Radicals

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

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

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

$\begin{array}{cccccc}\text{The radical terms are isolated.}\hfill & & & \hfill \sqrt[3]{4x-3}& =\hfill & \sqrt[3]{3x+2}\hfill \\ \begin{array}{c}\text{Since the index is 3, cube both sides of the}\hfill \\ \text{equation.}\hfill \end{array}\hfill & & & \hfill {\left(\sqrt[3]{4x-3}\right)}^{3}& =\hfill & {\left(\sqrt[3]{3x+2}\right)}^{3}\hfill \\ \text{Simplify, then solve the new equation.}\hfill & & & \hfill 4x-3& =\hfill & 3x+2\hfill \\ & & & \hfill x-3& =\hfill & 2\hfill \\ & & & \hfill x& =\hfill & 5\hfill \\ & & & \hfill \text{The solution is}\phantom{\rule{0.2em}{0ex}}x& =\hfill & 5.\hfill \\ \text{Check the answer.}\hfill & & & & & \\ \text{We leave it to you to show that 5 checks!}\hfill & & & & & \end{array}$

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

$x=3$

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

$x=-\frac{6}{5}$

Sometimes after raising both sides of an equation to a ability, we withal have a variable inside a radical. When that happens, we repeat Pace ane and Pace ii of our procedure. We isolate the radical and raise both sides of the equation to the ability of the index again.

How to Solve a Radical Equation

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

$x=4$

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

$x=9$

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

Solve a radical equation.

Isolate one of the radical terms on i side of the equation.
Heighten both sides of the equation to the ability of the index.
Are at that place any more radicals?
If yep, repeat Step 1 and Stride two again.

If no, solve the new equation.
Check the answer in the original equation.

Be careful every bit y'all square binomials in the next example. Call up the pattern is ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$ or ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}.$

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


The radical on the right is isolated. Square both sides.
Simplify.
In that location is still a radical in the equation so we must repeat the previous steps. Isolate the radical.
Foursquare both sides. It would not help to divide both sides by 6. Remember to square both the 6 and the $\sqrt{q-2}.$
Simplify, then solve the new equation.
Distribute.
It is a quadratic equation, so get cipher on ane side.
Cistron the right side.
Utilise the Zero Production Property.
The checks are left to you.	The solutions are $q=6$ and $q=2.$

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

$x=5$

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

$x=0\phantom{\rule{0.2em}{0ex}}x=4$

Utilise 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 the states a plan for solving applications with formulas from whatsoever subject area.

Utilize a trouble solving strategy for applications with formulas.

Read the problem and make sure all the words and ideas are understood. When appropriate, depict a figure and characterization information technology with the given information.
Place what we are looking for.
Name what we are looking for by choosing a variable to correspond it.
Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given data.
Solve the equation using good algebra techniques.
Cheque the respond in the problem and make sure it makes sense.
Answer the question with a complete sentence.

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

Falling Objects

On Earth, if an object is dropped from a tiptop of h feet, the fourth dimension in seconds it will have 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 anxiety, nosotros can notice the time it takes to attain the basis past substituting $h=64$ into the formula.



Take the square root of 64.
Simplify the fraction.

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

Marissa dropped her sunglasses from a bridge 400 anxiety above a river. Employ the formula $t=\frac{\sqrt{h}}{4}$ to find how many seconds it took for the sunglasses to achieve the river.

Step 1. Read the problem.
Pace ii. Identify what we are looking for.	the time information technology takes for the sunglasses to reach the river
Step 3. Name what we are looking.	Let $t=$ time.
Step four. Translate into an equation past writing the appropriate formula. Substitute in the given information.
Step 5. Solve the equation.

Step half-dozen. Cheque the respond in the problem and make sure it makes sense.
Does five seconds seem like a reasonable length of fourth dimension?	Yes.
Stride seven. Answer the question.	It volition take five seconds for the sunglasses to reach the river.

A helicopter dropped a rescue packet from a elevation of 1,296 feet. Apply the formula $t=\frac{\sqrt{h}}{4}$ to find how many seconds information technology took for the package to reach the ground.

9 seconds

A window washer dropped a squeegee from a platform 196 anxiety in a higher place the sidewalk Employ the formula $t=\frac{\sqrt{h}}{4}$ to find how many seconds it took for the squeegee to reach the sidewalk.

$3.5$ seconds

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

Sideslip Marks and Speed of a Motorcar

If the length of the skid marks is d feet, so the speed, s, of the car before the brakes were practical tin be found by using the formula

$s=\sqrt{24d}$

Later on a car accident, the slip marks for one motorcar measured 190 feet. Use the formula $s=\sqrt{24d}$ to find the speed of the car before the brakes were practical. Round your respond to the nearest 10th.

Step 1. Read the problem
Footstep 2. Identify what we are looking for.	the speed of a motorcar
Step iii. Name what weare looking for,	Allow $s=$ the speed.
Footstep 4. Translate into an equation by writing the appropriate formula. Substitute in the given data.
Pace five. Solve the equation.

Round to i decimal place.

	The speed of the car before the brakes were applied was 67.5 miles per hr.

An blow investigator measured the skid marks of the car. The length of the skid marks was 76 feet. Utilise the formula $s=\sqrt{24d}$ to observe the speed of the car before the brakes were practical. Circular your reply to the nearest tenth.

$42.7$ anxiety

The sideslip marks of a vehicle involved in an accident were 122 anxiety long. Use the formula $s=\sqrt{24d}$ to find the speed of the vehicle before the brakes were practical. Circular your answer to the nearest 10th.

$54.1$ feet

Key Concepts

Binomial Squares
$\begin{array}{c}{\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}$
Solve a Radical Equation
1. Isolate ane of the radical terms on one side of the equation.
2. Raise both sides of the equation to the power of the index.
3. Are there any more radicals?
  If yes, echo Step 1 and Footstep 2 once more.
  
  If no, solve the new equation.
4. Check the reply in the original equation.
Problem Solving Strategy for Applications with Formulas
1. Read the problem and make certain all the words and ideas are understood. When appropriate, draw a figure and label it with the given information.
2. Place what we are looking for.
3. Name what we are looking for by choosing a variable to stand for 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. Cheque the answer in the problem and brand sure it makes sense.
7. Answer the question with a consummate sentence.
Falling Objects
- On Earth, if an object is dropped from a top of h anxiety, the time in seconds information technology will take to reach the basis is found by using the formula $t=\frac{\sqrt{h}}{4}.$
Slip Marks and Speed of a Car
- If the length of the skid marks is d feet, then the speed, due south, of the car before the brakes were applied can be found past using the formula $s=\sqrt{24d}.$

Do Makes Perfect

Solve Radical Equations

In the following exercises, solve.

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

$m=14$

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

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

no solution

$\sqrt{3y-4}=-2$

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

$x=-4$

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

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

$m=14$

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

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

$v=17$

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

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

$m=\frac{7}{2}$

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

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

no solution

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

$\sqrt{u-3}-3=u$

$u=3,u=4$

$\sqrt{v-10}+10=v$

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

$r=1,r=2$

$\sqrt{s-8}=s-8$

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

$x=10$

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

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

$x=-8$

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

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

$x=8$

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

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

$x=-4$

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

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

$x=7$

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

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

$x=3$

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

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

$z=21$

$\sqrt{w+25}-w=-5$

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

$x=42$

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

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

$r=3$

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

Solve Radical Equations with 2 Radicals

In the following exercises, solve.

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

$u=3$

$\sqrt{4v+1}=\sqrt{3v+3}$

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

$r=-2$

$\sqrt{10+2c}=\sqrt{4c+16}$

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

$x=1$

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

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

$x=-8,x=2$

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

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

$a=0$

$\sqrt{r}+6=\sqrt{r+8}$

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

$u=\frac{9}{4}$

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

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

$a=4$

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

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

$x=0\phantom{\rule{0.2em}{0ex}}x=4$

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

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

$x=1\phantom{\rule{0.2em}{0ex}}x=5$

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

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

$x=9$

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

Use Radicals in Applications

In the post-obit exercises, solve. Round approximations to one decimal place.

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

$8.7$ anxiety

Landscaping Vince wants to make a square patio in his one thousand. He has enough concrete to pave an surface area of 130 foursquare anxiety. Utilize the formula $s=\sqrt{A}$ to detect the length of each side of his patio. Round your answer to the nearest tenth of a pes.

Gravity A hang glider dropped his jail cell telephone from a elevation of 350 feet. Use the formula $t=\frac{\sqrt{h}}{4}$ to discover how many seconds information technology took for the prison cell phone to accomplish the ground.

$4.7$ seconds

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

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

72 feet

Blow investigation An accident investigator measured the skid marks of one of the vehicles involved in an blow. 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 practical. Round your reply to the nearest tenth.