Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 2

Algebra 2 Solving Radical Equations and Inequalities

Name: Algebra 2: Solving Radical Equations and Inequalities
Brand: Educator.com
Price: 35 USD
Availability: InStock

Section 7: Radical Expressions and Inequalities: Lecture 7 | 31:27 min

Lecture Description

Next Lecture

Previous Lecture

Discussion
Answer Engine
Study Guides
Practice Questions
Download Lecture Slides
Table of Contents
Transcription
Related Books

Lecture Comments (4)

1 answer

Last reply by: Dr Carleen Eaton
Sun Jul 8, 2018 1:04 PM

Post by Okwudili Ezeh on June 27, 2018

In example 4, why are you writing b as 2 rather than -2.

1 answer

Last reply by: Shiden Yemane
Sat Feb 10, 2018 11:04 AM

Post by Shiden Yemane on February 10, 2018

On example 4, you wrote the radicand to be x+2. Shouldn't it be x+21?
Thanks

Answer EngineGet answers to any question!Ask any question related to Algebra 2

Working on the solution...

Solving Radical Equations and Inequalities

To solve a radical equation, raise both sides of the equation to the power equal to the index of the radical. This will eliminate at least one radical. If a radical remains in the new equation, repeat the process.
Always check for extraneous solutions – values for which one or more radicals in the original equation have a negative radicand. Exclude such values from the solution set.

Solving Radical Equations and Inequalities

Solve − 10 + √{v + 6} = − 5

Step 1 - Isolate the radical on one side of the equation
− 10 + √{v + 6} = − 5 = >√{v + 6} = 5
Step 2 - Raise both sides to the second power to eliminate the square root
( √{v + 6} )² = 5²
v + 6 = 25
Step 3 - Solve
v = 19
Step 4 = Check for extraneous solutions.
√{v + 6} = 5
√{19 + 6} = 5
√{25} = 5
5 = 5

Solution is not an extraneous solution, v = 19

Solve 0 = − 6 + √{35p + 1}

Step 1 - Isolate the radical on one side of the equation
0 = − 6 + √{35p + 1} = > 6 = √{35p + 1}
Step 2 - Raise both sides to the second power to eliminate the square root
6² = ( √{35p + 1} )²
36 = 35p + 1
Step 3 - Solve
35 = 35p
p = 1
Step 4 = Check for extraneous solutions.
6 = √{35p + 1}
6 = √{35(1) + 1}
6 = √{36}
6 = 6

Solution is not an extraneous solution, p = 1

Solve 0 = − 6 + √{[p/10]}

Step 1 - Isolate the radical on one side of the equation
0 = − 6 + √{[p/10]} = > 6 = √{[p/10]}
Step 2 - Raise both sides to the second power to eliminate the square root
6² = ( √{[p/10]} )²
36 = [p/10]
Step 3 - Solve
36(10) = [p/10]10 = > 36(10) = [p/]
p = 360
Step 4 = Check for extraneous solutions.
6 = √{[p/10]}
6 = √{[360/10]}
6 = √{36}
6 = 6

Solution is not an extraneous solution, p = 360

Solve − 1 = √{5x − 1} − √{8x + 1}

Step 1 - Isolate one radical to the left side to make solving easier.
− 1 = √{5x − 1} − √{8x + 1} → √{8x + 1} − 1 = √{5x − 1}
Step 2 - Raise both sides to the second power to eliminate the square roots. Multiply using FOIL or
any other method you prefer.
( √{8x + 1} − 1 )² = ( √{5x − 1} )²
( √{8x + 1} − 1 )( √{8x + 1} − 1 ) = 5x − 1
Multiply on the left sde of the equation
√{8x + 1} -1
√{8x + 1} 8x + 1 − √{8x + 1}
-1 − √{8x + 1} + 1
8x + 2 − 2√{8x + 1} = 5x − 1
Step 3 - Isolate radical to one side of the equation and raise both sides to the second power.
3x + 3 = 2√{8x + 1}
( 3x + 3 )² = ( 2√{8x + 1} )²
9x² + 18x + 9 = 4(8x + 1)
9x² + 18x + 9 = 32x + 4
9x² − 14x + 5 = 0
Step 4 = Solve by factoring
9x² − 14x + 5 = (9x − 5)(x − 1) = 0
x = [5/9] and x = 1
Step 5 - Check Solutions
x=[5/9]
− 1 = √{5x − 1} − √{8x + 1}
− 1 = √{5[5/9] − 1} − √{8[5/9] + 1}
− 1 = √{[25/9] − 1} − √{[40/9] + 1}
− 1 = √{[16/9]} − √{[49/9]}
− 1 = √{[16/9]} − √{[49/9]}
− 1 = [4/3] − [7/3] = − [3/3] = − 1
x=1
− 1 = √{5x − 1} − √{8x + 1}
− 1 = √{5(1) − 1} − √{8(1) + 1}
− 1 = √4 − √9
− 1 = 2 − 3 = − 1

x = [5/9] and x = 1

Solve 1 = √{5 − 4x} − √{6 − 2x}

Step 1 - Isolate one radical to the left side to make solving easier.
1 = √{5 − 4x} − √{6 − 2x} → 1 + √{6 − 2x} = √{5 − 4x}
Step 2 - Raise both sides to the second power to eliminate the square roots. Multiply using FOIL or
any other method you prefer.
( 1 + √{6 − 2x} )² = ( √{5 − 4x} )²
( 1 + √{6 − 2x} )( 1 + √{6 − 2x} ) = 5 − 4x
Multiply on the left side of the equation
√{6 − 2x} 1
1 √{6 − 2x} 1
√{6 − 2x} 6 − 2x √{6 − 2x}
7 − 2x + 2√{6 − 2x} = 5 − 4x
Step 3 - Isolate radical to one side of the equation and raise both sides to the second power.
2√{6 − 2x} = − 2x − 2
( 2√{6 − 2x} )² = ( − 2x − 2 )²
4(6 − 2x) = 4x² + 8x + 4
24 − 8x = 4x² + 8x + 4
4x² + 16x − 20 = 0
Step 4 = Solve by factoring, divide everything by 4.
[(4x² + 16x − 20 = 0)/4]
x² + 4x − 5 = (x + 5)(x − 1)
x = − 5 and x = 1
Step 5 - Check Solutions
x = −5
1 = √{5 − 4x} − √{6 − 2x}
1 = √{5 − 4( − 5)} − √{6 − 2( − 5)}
1 = √{25} − √{16}
1 = 5 − 4 = 1
Checks!
x=1
1 = √{5 − 4x} − √{6 − 2x}
1 = √{5 − 4(1)} − √{6 − 2(1)}
1 = √1 − √4
1 = 1 − 2 = − 1
Not a solution

x = − 5

Solve 3 = √{ − 2 − 9x} − √{ − 1 − x}

Step 1 - Isolate one radical to the left side to make solving easier.
3 = √{ − 2 − 9x} − √{ − 1 − x} = > 3 + √{ − 1 − x} = √{ − 2 − 9x}
Step 2 - Raise both sides to the second power to eliminate the square roots. Multiply using FOIL or
any other method you prefer.
( 3 + √{ − 1 − x} )² = ( √{ − 2 − 9x} )²
( 3 + √{ − 1 − x} )( 3 + √{ − 1 − x} ) = − 2 − 9x
Multiply on the left sde of the equation
3 √{ − 1 − x}
3 9 3√{ − 1 − x}
√{ − 1 − x} 3√{ − 1 − x} − 1 − x
8 − x + 6√{ − 1 − x} = − 2 − 9x
Step 3 - Isolate radical to one side of the equation and raise both sides to the second power.
6√{ − 1 − x} = − 8x − 10
( 6√{ − 1 − x} )² = ( − 8x − 10 )²
36( − 1 − x) = 64x² + 160x + 100
− 36 − 36x = 64x² + 160x + 100
64x² + 196x + 136 = 0
Step 4 = Solve by factoring, divide everything by 4.
[(64x² + 196x + 136 = 0)/4]
16x² + 49x + 34 = (16x + 17)(x + 2)
x = − [17/16] and x = − 2
Step 5 - Check Solutions
x=−[17/16]
3 = √{ − 2 − 9x} − √{ − 1 − x}
3 = √{ − 2 − 9( − [17/16] )} − √{ − 1 − ( − [17/16] )}
3 = √{ − 2 − 9( − [17/16] )} − √{ − 1 − ( − [17/16] )}
3 = √{ − 2 + ( [153/16] )} − √{ − 1 + ( [17/16] )}
3 = √{[121/16]} − √{[1/16]} = [11/4] − [1/4] = [10/4] = [5/2]
Does Not Check!
x=−2
3 = √{ − 2 − 9x} − √{ − 1 − x}
3 = √{ − 2 − 9( − 2)} − √{ − 1 − ( − 2)}
3 = √{ − 2 + 18} − √{ − 1 + 2}
3 = √{16} − √1 = 4 − 1 = 3
Checks! A solution

x = − 2

Solve 7 + √{x + 7} ≤ 14

Step 1: Isolate the square root on one side of the inequality
√{x + 7} ≤ 7
Step 2: Determine the domain of the inequality. You cannot have negative inside the radical.
x + 70
x − 7
Step 3: Raise both sides to the second power
( √{x + 7} )² ≤ ( 7 )²
x + 7 ≤ 49
x ≤ 41
Step 4: Combine the restrictions of the domain with previous answer

− 7 ≤ x ≤ 41

Solve √{ − 8 − 12x} − 5 ≤ 5

Step 1: Isolate the square root on one side of the inequality
√{ − 8 − 12x} ≤ 10
Step 2: Determine the domain of the inequality. You cannot have negative inside the radical.
− 8 − 12x0
x ≤ − [8/12]
x ≤ − [2/3]
Step 3: Raise both sides to the second power
( √{ − 8 − 12x} )² ≤ ( 10 )²
− 8 − 12x ≤ 100
− 12x ≤ 108
x − 9
Step 4: Combine the restrictions of the domain with previous answer

− 9 ≤ x ≤ − [2/3]

Solve √{3 − 39x} − 9 ≤ 0

Step 1: Isolate the square root on one side of the inequality
√{3 − 39x} ≤ 9
Step 2: Determine the domain of the inequality. You cannot have negative inside the radical.
3 − 39x0
− 39x − 3
x ≤ − [3/( − 39)] = [1/13]
Step 3: Raise both sides to the second power
( √{3 − 39x} )² ≤ ( 9 )²
3 − 39x ≤ 81
− 39x ≤ 78
x − 2
Step 4: Combine the restrictions of the domain with previous answer

− 2 ≤ x ≤ [1/13]

Solve 9√{12x − 3} − 81 ≤ 0

Step 1: Isolate the square root on one side of the inequality
9√{12x − 3} ≤ 81
Step 2: Determine the domain of the inequality. You cannot have negative inside the radical.
12x − 3 ≥ 0
12x ≥ 3
x[3/12] = [1/4]
Step 3: Raise both sides to the second power, but divide both sides by 9 first.
[(9√{12x − 3} )/9] ≤ [81/9]
( √{12x − 3} )² ≤ ( 9 )²
12x − 3 ≤ 81
12x ≤ 84
x ≤ 7
Step 4: Combine the restrictions of the domain with previous answer

[1/4] ≤ x ≤ 7

« Previous Question | Next Question »

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Solving Radical Equations and Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Radical Equations

Extraneous Roots

Eliminating nth Roots

Radical Inequalities

Example 1: Solve Radical Equation

Example 2: Solve Radical Equation

Example 3: Solve Radical Inequality

Example 4: Solve Radical Equation

Intro 0:00
Radical Equations 0:11

Variables in Radicands
Example: Radical Equation
Example: Complex Equation

Extraneous Roots 7:21

Squaring Technique
Double Check
Example: Extraneous

Eliminating nth Roots 10:04

Isolate and Raise Power
Example: nth Root

Radical Inequalities 11:27

Restriction: Index is Even
Example: Radical Inequality

Example 1: Solve Radical Equation 15:41
Example 2: Solve Radical Equation 17:44
Example 3: Solve Radical Inequality 20:24
Example 4: Solve Radical Equation 24:34

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Solving Radical Equations and Inequalities

Welcome to Educator.com.0000

In previous lessons, we have learned techniques for working with radicals.0003

Now, we are going to put those together and use them to solve radical equations and inequalities.0007

First, the definition of a radical equation: recall that radical equations contain radicals with variables in the radicand.0012

For example, it is something like the fourth root of x + 5 equals 3, or 6 + √(2x + 1) = 4.0020

These are both examples of radical equations.0033

If you had a radical, and there was just a number under it, like the square root of 5, plus 2x,0036

this is not a radical equation, because there is no variable under the radical sign.0047

OK, in order to solve these equations, we are going to use a technique learned earlier on in Algebra I and reviewed now, and also go into more complex problems.0056

For example, if we had the square root of x + 2, minus 6, equals -1; what we are going to do is0066

raise each side of the equation to a power that is the index of the radical.0077

So, the index of the radical here is 2; so I am going to raise both sides to the second power; in other words, I am going to square both sides.0083

But before you do that, first isolate the radical; then raise both sides to a power equal to the index of the radical.0090

First, I am isolating this by adding 6 to both sides; so, -1 + 6 is going to give me 5.0110

Now, I am going to square both sides; I am going to raise both sides to that index power, which is 2.0120

Well, since the index is 2, and it is being raised to the second power, this equals the radicand, and this equals 25.0126

Subtracting 2 from both sides gives me x = 23.0136

So, to solve radical equations, you isolate the radical, then raise both sides to a power equal to the value of the index; then, solve the equation.0140

Now, you see here that it says sometimes this must be done twice in order to eliminate all radicals.0152

That is a more complex situation that we are going to look at right now.0158

Let's say I am given √(x + 3) - 2 = √(x - 5).0163

I can't just isolate the radical, because there is more than one radical.0171

So in this case, I just begin by raising both sides to a power that is equal to the index.0175

Since the index is 2, I am just going to first square both sides; I can't isolate.0183

Squaring both sides is going to give me this; and if I look at this left side, I am squaring a difference; that is analogous to this situation:0193

(a - b)², which turns out to be a² - 2ab + b².0209

I just substitute in here; a is the square root of x + 3, and b is 2.0215

So, that makes my work on the left much quicker, instead of having to use FOIL, use the distributive property,0222

multiply everything out, add...I just say, "OK, the square root of x + 3, squared, minus 2 times a times b..."0227

a is the square root of x + 3, and then b is 2.0239

And then, finally, I get b².0248

On the right, I raise this to the index power, so I get the radicand, x - 5.0256

OK, I squared the square root of x + 3; the square root of (x + 3)² is simply the radicand, x + 3.0262

Here, I have 2 times 2 is 4, times the square root of x + 3.0272

Here I have -2 times -2 is 4.0279

Now, you can see that the radical on the right was eliminated.0285

However, I still have a radical on the left; so that is why I am going to need to repeat this process.0288

I squared both sides, and then I need to repeat if radicals remain.0293

I squared both sides, and I am going to need to repeat; let's simplify first.0305

This is 3 + 4, so this gives me x + 7, minus 4√(x + 3), equals x - 5.0309

I am going to move this 7 over to the right by subtracting 7 from both sides; and that is going to be x - 5 - 7, so x - 12.0320

Now, the next thing is: to isolate this, I want to subtract x from each side; so x - x...that drops out.0332

x -x...that drops out, too: conveniently, the x's dropped out, making this even simpler to work with.0339

I still want to isolate this completely; so, to do that, I need to divide both sides by -4.0351

Now, I am left with something easier to work with, because I end up with just the radical isolated on the left, and then I just have a number on the right.0365

So, I am going to repeat my process, and I am going to square this; and let's start right up here with that second process, √(x + 3)² = 3².0376

I raised this radical to the index power; that is going to leave me with the radicand, x + 3; and on the right, I have 9.0395

This gives me x = 6; this was pretty complex.0404

And I solved it by first squaring both sides, which got rid of the radical on the right.0407

Then, I simplified the left, found that the x's dropped out, combined my constants, and ended up with this.0412

To completely isolate the radical, I divided by 4; and when I ended up with this, then I have something that is more familiar,0424

where I have an isolated radical expression on the left, and squared both sides, solved, and determined that x equals 6.0431

OK, recall that sometimes, when we use this method of squaring both sides, we end up with a solution;0442

and when we check the solution, it actually does not satisfy the original equation.0449

This is a result of the squaring step of using this technique; and these solutions are called extraneous solutions.0456

Therefore, you always must check the solutions you get from this technique back in the original equation or inequality.0464

If the solution exists, and you use this technique, the solution will be among your results.0474

And I say "among your results" because you might have the valid solution, plus some extraneous solutions.0478

There may be no solution; you may just end up with one or two extraneous solutions; you may end up with one or more valid solutions.0485

There is no way of knowing; but if the solution exists, it will be somewhere among the solutions you end up with.0492

The only way to find it, though, is to check all the solutions to see what you have.0497

For example, we did, earlier on, that example (which was a little bit complicated): √(x + 3) - 2 = √(x - 5).0503

And when we worked that out, I came out with the solution x = 6.0521

I can't just assume it is a valid solution; it actually may not be valid.0525

The only way to know is to go back and substitute in.0529

Everywhere there is an x, I am going to make it a 6; and I am going to see if my equation holds up.0534

This is going to give me √(x + 3), which is √9, minus 2, equals √(6 - 5); so that becomes √1.0542

The square root of 9 is 3; minus 2 equals the square root of 1, which is 1.0557

1 does equal 1; this is true, so the solution, which is x = 6, is valid.0564

It is possible sometimes that you will get a solution, and you check it, and you get something strange like 4 = 5, which is not true.0574

That means that this answer, this solution, does not satisfy the equation; and therefore, it was an extraneous solution0583

that popped up as a result of squaring both sides.0591

I could have ended up with two solutions or three solutions; and some were extraneous/some were valid/none were valid.0596

Again, the only way to tell is to check.0601

Eliminating n^th roots: we have been doing this, but just to bring it out and explain exactly what we were doing:0605

to eliminate an n^th root, you isolate the root (I talked about isolating the radicals),0611

and then raise both sides of the equation to the power n.0618

That is what we did when we took something like √(x + 2) - 4 = 8, and we first isolated, which would give me 12.0622

And then, I said, "OK, n is 2, so I am going to square it; I am going to raise it to the second power."0638

So, this is exactly what we have been doing: isolate the root, and then raise both sides of the equation to that index power.0649

And the same technique could apply to if you are using a higher index; but it does become more complicated.0657

OK, and again, an important last step is to check solutions in the original equation to make sure that they are valid.0666

They may be valid; they may be extraneous.0683

All right, we talked about radical equations; and radical inequalities are very similar.0687

So, these are inequalities with variables in the radicand.0693

And just as we talked about with equations (radical equations have variables in the radicand), radical inequalities0697

are the same thing, only we are working with greater than or less than, or greater than or equal to, and all that, instead of equals.0703

We have this restriction that we have seen before with radicals: if the index is even,0713

we have to determine the values of the variable for which the radicand is greater than or equal to 0.0717

In other words, we have to restrict the solution set to values that will not make the radicand negative,0723

because if the radicand is negative, and we have an even index, we will end up with something that is not a real number.0732

So, after we determined that restriction, we solved algebraically, using the same techniques that we have talked about before with radicals.0739

Let's use an example: the square root of 3x + 4, minus 2, is greater than 5.0747

Before I even begin trying to solve this, I am going to figure out what my excluded values are.0754

So, I know that 3x + 4 has to be greater than or equal to 0.0760

If this ended up being negative, and I have an even index here, 2, then I can end up with a complex number as a solution.0765

And I don't want that; we are just working with real numbers.0773

So, I am going to go ahead and say 3x, and I am going to subtract 4 from both sides to get 3x ≥ -4.0777

And then, I am going to divide both sides by 3; so x ≥ -4/3.0785

This is just saying that I am restricting the domain to these values.0792

Any value of x that is less than -4/3 is an excluded value.0797

So, x less than -4/3 are excluded.0803

I haven't even begun to solve this yet; I was just figuring out what the restrictions are on my solution set.0810

OK, so now, I am going to go ahead and actually solve this algebraically, using techniques that we used with radical equations.0818

The first one here is to isolate the radical on one side of this inequality.0828

And I am going to do that by adding 2 to both sides.0834

So, this is going to give me √(3x + 4) > 7.0843

Next, I am going to raise both sides to the power equal to the index, which, in this case, would be the power 2.0852

So, squaring both sides is going to give me...the square root squared is going to give me the radicand,, 3x + 4 > 7 squared, or 49.0860

Now, I just solve: 3x...subtracting 4 from both sides, the 4 drops out from here...49 - 4 is 45.0875

3x > 45; and I want to isolate the x, so I am going to divide by 3; 3 is going to drop out--I get x > 45/3, or x>15.0885

And I check and say, "OK, is this allowed?"0901

Yes, it is, because if x is greater than 15, it is going to be greater than -4/3.0904

So, I am not going into values that are not allowed.0912

And then, you could check this by picking a value greater than 15,0916

and putting it in here, back into the original, and making sure that the inequality is true--that it holds up.0921

I could pick a value such as 20 or 16, put it in here, solve this, and make sure that I ended up with something greater than 5,0927

because again, when we use this technique of squaring both sides, we need to check the answers.0936

OK, first example: I am going to solve this radical equation by the technique of first isolating the radical.0942

So, all I have to do in order to achieve that is subtract 8 from both sides to get √(3x - 2) = 7.0957

Then, I am going to square both sides, because the index is 2.0965

So, if I raise both sides to the second power, then I am going to get the radicand from here,0969

because the square root of 3x - 2, times itself, is 3x - 2.0976

On the right, I get 7², is 49.0983

Now, I just solve using my usual algebraic techniques, adding 2 to both sides.0986

3x = 51; now, dividing both sides by 3, x = 51/3; and if you do the arithmetic on that, you will find that x equals 17.0993

Another important step is checking to make sure this is a valid solution.1007

So, I am going to go all the way back to this original; and I am going to let x equal 171010

and substitute that in to get 8 + √3 times 17 minus 2 equals 15.1017

3 times 17 is 51; 51 minus 2 is 49; so, this is going to give me 8 + √49 (which is 7) = 15.1028

Since 15 = 15, then x = 17 is valid, because x = 17 actually satisfied this equation--the two sides were equal.1048

OK, the second example is a similar idea; but instead of working with square roots, now the index is 4.1065

But I am going to use the same technique: I am going to first isolate this radical.1072

And I am going to do that by subtracting 4 from both sides.1077

6 - 4 is 2, so now I have that the fourth root of 2x - 8 equals 2.1086

In this case, since the index power is 4, I am going to raise both sides to the fourth power.1094

OK, I have the fourth root of 2x - 8 raised to the fourth power.1104

If the index and the power you are raising it to are the same, this, then, equals the radicand.1110

2 to the fourth power is...2 times 2 is 4, times 2 is 8, times 2 is 16.1119

Next, I just isolate the x, using algebraic techniques: I am going to add 8 to both sides--that gives me 2x = 24.1129

Dividing both sides by 2 gives me x = 12.1140

And we can't forget the important step of checking our solution back in the original equation.1144

I am going to check, and this is the fourth root of 2...I am going to check x = 12.1150

The fourth root of 2 times 12, minus 8, plus 4, equals 6; let's see if that holds up.1162

2 times 12 is 24, minus 8, plus 4, equals 6.1173

This is going to give me the fourth root of 24 - 8, which is 16.1180

The fourth root of 16 is 2, because 2⁴ = 16; so the fourth root of 16 is 2.1187

That leaves me with 2 + 4 = 6; and 6 does equal 6, so x = 12 is valid.1202

It is a valid solution, because it satisfies this equation.1212

When x is 12, the equation holds true.1216

In our third example, we have an inequality: we have to take that extra step of excluding values--finding what the excluded values are.1224

3x - 6 must be greater than or equal to 0; it needs to be a non-negative number,1235

because since this is an even power, if this radical is the square root of a negative number,1242

then I could end up with an imaginary number, which I don't want.1251

So, since this is an even index, I am going to take the extra step and define the values of x that I am allowed to work with.1255

OK, 3x - 6 must be greater than or equal to 0; adding 6 to both sides gives me 3x ≥ 6.1263

Now, dividing both sides by 3, x ≥ 2.1275

So, I have not solved it yet; all I have said is that the only values that are even allowable to use are these.1280

Now, I am going to go about solving it; so let's rewrite this here.1287

2 plus the square root of 3x minus 6 is less than or equal to 8.1291

Just like with radical equations, I start out by isolating the radical.1296

So, I am going to subtract 2 from both sides; my index value is 2--it is just the square root.1301

So, I am going to square both sides.1309

The square root of something squared is just that value; so this becomes the radicand, because the square root of 3x - 6, times itself, is 3x - 6.1317

It is less than or equal to 36.1331

Now, I am going to add 6 to both sides: 36 + 6 is 42.1334

Divide both sides by 3: 42 divided by 3 is actually 14, so x is less than or equal to 14.1343

I am not quite done: I need to put all of this together, because if I just said,1355

"OK, x is less than or equal to 14," I might say, "Oh, OK, x can be 0," but it actually can't,1359

because I have this restriction right here on the domain that it has to be greater than or equal to 2.1365

Therefore, I am going to write out the whole thing together, saying that x needs to be greater than or equal to 21370

in order to even be a value that will give me a real number.1380

But to actually solve this inequality, to find the solution set...the solution set only encompasses those values of x that are less than or equal to 14.1386

So, in this case, I actually had to put this all together.1394

Now, in order to check this, one way to check it is to choose a value that is within the solution set (for example, 5--1400

5 is greater than or equal to 2, and less than or equal to 14).1408

So, I am going to check with x = 5, and I am going to see if that satisfies this inequality.1412

2 + √(3x - 6) ≤ 8: if I am going to let x equal 5, and just make sure that this thing holds up,1420

this is going to give me 15 - 6 under the radical, which is...15 - 6 is 9, so I have the square root of 9 there,1434

which is 3; so I end up with 5 ≤ 8, and this is true.1448

So, this helped to verify that I actually have a valid solution.1457

Again, when you are using this technique of squaring both sides, extraneous solutions can occur.1466

OK, this time, I actually have two radicals; so it is not going to be possible to just isolate the radical.1474

Therefore, my first step is just going to be to get rid of at least one of the radicals by squaring both sides.1481

So, I am going to go ahead and square both sides, and recall that, on the left, if I have something1489

such (a - b)², it is going to give me a² - 2ab + b².1501

And by remembering that, I will save myself the work of having to use the distributive property and figure this whole thing out.1507

I am just going to go ahead and say, "OK, a equals the square root of x plus 21, and b equals 2."1514

So, on the left, I am going to get a²...the square root of x plus 21, squared...minus 2 times a1523

(the square root of x plus 21), times b (which is 2); and then, for the last term, I will get b², which is 2².1537

On the right, I have a -1 in front of this; when I square that, -1 times -1 becomes 1.1551

So, I am going to end up with a positive expression on the right, and the square root of x + 5 squared...1558

since the index is the same as the power, this becomes the radicand; so on the right, I end up with x + 5.1568

OK, simplifying: over here, I have a square root raised to the second power; the square root squared gives me the radicand.1576

So, I can see I have gotten rid of this radical on the right; and this is not a radical.1586

However, in the middle, this middle term is -2, times 2 is -4, times the square root of x plus 21.1592

I am still left with this radical; here, 2 times 2 is 4; simplify before you proceed.1600

This gives me 2 and 4, so that is x + 6 minus this equals x + 5.1612

So, I am going to go ahead and subtract this 6 from both sides...this is actually 21.1622

So, go ahead and simplify this; this is 21 plus 4, is actually going to be 25.1642

So, x + 21, the square root of that squared, gives me x + 21; so that is 25 right here; 21 + 4 is 25.1649

OK, simplify by subtracting 25 from both sides: 5 - 25 is -20.1662

In the next step, I would subtract x from each side; so when I do that, I get x - x; the x drops out of the left.1675

When I take x - x, the x also drops out of the right side.1687

Now, I have gotten rid of this radical on the right; on the left, I have a radical,1693

but I also have a -4 in front of it, so I need to isolate the radical by dividing both sides by -4.1698

So, this gives me -20 divided by -4 on the right; so the square root of x + 21 equals 5.1706

Since I still have a radical left, I have to go through this process again.1717

So, let's take this up here and repeat the process of squaring both sides.1721

The square root of x + 21, squared, gives me the radicand: x + 21 = 5²...that is 25.1734

Subtracting 21 from both sides gives me 4; so I came up with this solution that x = 4, and I need to check that in the original.1743

So, check by inserting 4 for each x in the original.1755

That is going to give me the square root of 4 + 21, minus 2, equals negative...and then that is the square root of 4 + 5.1763

This gives me the square root of 25, minus 2, equals minus the square root of 9.1776

The square root of 25 is 5, minus 2 equals minus the square root of 9, which is 3; so that gives me -3.1783

This is 5 - 2 = -3; this is not true; therefore, the solution is not valid.1795

This solution, x = 4, is not valid.1807

Now, I said that, when you use this method, if there is a valid solution, you will come up with it.1809

You might have some extra solutions, but you will have a valid solution, if it exists.1815

What this tells me, since this is not valid, is that there is no valid solution here.1820

So, this was a pretty complex problem: there were two radicals, so we had to square to get rid of the radical on the right;1832

do a bunch of simplifying; isolate the remaining radical, which is on the left;1840

and repeat the process with that by squaring both sides to come up with x = 4.1847

And then, after all that work, we went and checked it, and found that x = 4 does not satisfy this equation--1853

that, when we use that, we end up with 3 = -3.1860

Had that negative not been there, we could have come up with a valid solution.1865

But with that negative in the original, the solution was not valid.1871

Therefore, there is no solution to this radical equation.1875

That concludes this lesson on radical equations and inequalities.1880

Thanks for visiting Educator.com!1885

Related Books

Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

Publisher: McDougal Littell

Year: 1999

This classic textbook does a great job of explaining concepts and includes lots of in-depth, worked out examples in every section as well as practice problems.

Related Books

Name	Description	Link
BookRenter.com	BookRenter.com is simply the most reliable online textbook rental service.	Visit BookRenter.com
PhysicsForums.com Homework Help	Physics Forums is a scientific community for students looking for math & science help.	Visit PhysicsForums.com Homework Help

Dr. Carleen Eaton

Solving Radical Equations and Inequalities

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro