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 â€“ 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
 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} = 5^{2}
 v + 6 = 25
 Step 3  Solve
 v = 19
 Step 4 = Check for extraneous solutions.
 √{v + 6} = 5
 √{19 + 6} = 5
 √{25} = 5
 5 = 5
 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
 6^{2} = ( √{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
 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
 6^{2} = ( √{[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
 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}
 9x^{2} + 18x + 9 = 4(8x + 1)
 9x^{2} + 18x + 9 = 32x + 4
 9x^{2} − 14x + 5 = 0
 Step 4 = Solve by factoring
 9x^{2} − 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
 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) = 4x^{2} + 8x + 4
 24 − 8x = 4x^{2} + 8x + 4
 4x^{2} + 16x − 20 = 0
 Step 4 = Solve by factoring, divide everything by 4.
 [(4x^{2} + 16x − 20 = 0)/4]
 x^{2} + 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
 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) = 64x^{2} + 160x + 100
 − 36 − 36x = 64x^{2} + 160x + 100
 64x^{2} + 196x + 136 = 0
 Step 4 = Solve by factoring, divide everything by 4.
 [(64x^{2} + 196x + 136 = 0)/4]
 16x^{2} + 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
 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
 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
 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
 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
*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 screencaptured 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 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
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 a^{2}  2ab + b^{2}.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 b^{2}.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} = 3^{2}.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 outI 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 truethat 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 7^{2}, 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 equationthe 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 sidesthat 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 2^{4} = 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 valuesfinding what the excluded values are.1224
3x  6 must be greater than or equal to 0; it needs to be a nonnegative 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 2it 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, 51400
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 a^{2}  2ab + b^{2}.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 a^{2}...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 b^{2}, which is 2^{2}.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 = 5^{2}...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 equation1853
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
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