To solve equations with nested radicals like √(√a) = 3/2, convert radicals to fractional exponents (√a = a^(1/2), so √(√a) = a^(1/4)), combine exponents using the rule a^m × a^n = a^(m+n), then isolate the variable by raising both sides to the reciprocal of the exponent. For the equation √(√a) = 3/2, this yields a^(1/4) = 3/2, so a = (3/2)^(4/3) = 81/16^(1/3), which is the cube root of 81/16.
Deep Dive
Voraussetzung
- Keine Daten verfügbar.
Nächste Schritte
- Keine Daten verfügbar.
Deep Dive
Olympiad Mathematics | Indian | Can you solve this?Hinzugefügt:
Okay, let's solve this very very quickly. Right? So we move solution.
This is two roo<unk> a roo<unk> a equ= 3.
Now the first the the first step you should take is to remove this two from here very quickly. So to do that we divide by two then we divide by two. two will take itself out. Now the square root of a square root of a is on the left hand side and is equal to 3 /2. Okay, I think this is the first step we should take.
Then the next step we're going to take is to remove the square root from here one by one. We having two square signs.
So we're going to remove them one by one. And to remove the first one, we have the square root of a square root of a. This is going to be squared. Now, this is the first time we're squaring.
So, we are removing the first square root. And we have to square the other side as well.
So, what do we do? This one and this one are going to go. So we have a a [snorts] to be equal to 3 / 2^ 2 will just be um 9 / 4. 3 2 is 9. 2 is 4. Now we still have a root. Take note we still have a root. So we need to remove the root.
And to do that we'll have to square both sides of the equation again. So this will be the second time we are doing that. We have a roo<unk> a all squared and it's = 9 / 4 all².
Now from this point remember that this can be split. Please take note of this.
If you have a okay let me use another letter. If you have m n to the power of c. Okay, if you have this, you can express this as m to the power c * m to the power * n to the power of c.
You can do this.
Yes, you can. So that's what we're going to do over there in case you do not understand. And in the same way if you have um m / n to the power of c this can be you know split to give you m ^ c / n to power c. So this one will be applicable to this one. Right? Good. So let me remove this again.
And from here now we have our a 2 * the square root of a squ and here we have 9 squar this is 9 / 4 2. Okay. So I just want to show all the steps so that you not have any excuse not to understand. So here we have a squ multiply by this can take this out. So we have a then here 9 * 8 sorry 9 * 9 is 81 4 * 4 is 16 and at this point we multiply the left hand side and that will give a to the power of 3 cuz there's an invisible power of 1. So a to the^ 3 is equal to 81 over over 16. And what are we looking for? We are looking for the value of a.
And to remove this power of three, we have to multiply the power by its own reciprocal. So we have a ^ 3 * 1 / 3 and it's equal to 81 / 16. Now raised to the power of the same 1 / 3. I think it's um better for us this way. So this one takes this out for us and we have the value of a already. So our a is 81 / 16 to the power of 1 / 3. Now for those of you that do not know what power of 1 / 3 is it means the cube root. So the question is asking us to find the cube root of 81 / 16. And we can do that.
Okay. Let's see how we do that.
Okay. So, we are looking for the cube roots, right?
But obviously um the both of them are not perfect cubes. Yes, they are not perfect cubes.
So, we can stop at a being equal to the fourth root.
Okay. The the third root. Yes, that's the cube root. The cube root of 81 / 16.
So this is our value of a. But I would still like to apply another method to solve this because I believe it's going to be faster using the other method.
Remember the equation is 2<unk> a <unk>= 3 right and we want to solve this using another approach. So what we'll do like we did before took eight I mean two to the other side. So you just have roo<unk> a roo<unk> a = 3 / 2. So we'll leave this one the way it is. We now focus on the left. Remember a is under one square root sign. The first one. So we have a ^ 1 / 2. But then the second a is under two square root signs. So we have a to the^ 1 / 2 * 1 / 2 and on the other hand 3 / 2. I hope this is going to be faster.
Okay, this is going to be faster. We have a to the^ 1 / 2 * a to the power of multiply the power we have 1 / 4. This is equal to 3 over 2. Now guess what I will do? Since we are multiplying the left and they have the same base, we can you know pick one and add the bars. We have a then we have 1 / 2 + 1 / 4. This is all equal to 3 / 2. And to go on we have our a. The LCM of the powers is 4.
Then 4 / 2 is 2 * 1 is 2 + 4 / 4 is 1 * 1 is 1 and this is equal to 3 / 2. Now to go on we shall have to go on we are going to have a to the^ 3 / 4 to be equal to 3 / 2. So our target is to remove the power and we will multiply the power by its own reciprocal.
Okay. So we have a to the^ 3 / 4 multiply by its own reciprocal which is 4 over 3. [snorts] Then the reciprocal is what will reflect on the right. 3 /2 will now be raised to the same 4 over 3 because this is the new inclusion.
Right? So from here three will go into three, four will go into four and we have only a a is on the left and is equal to a is equal to 3 / 2 to ^ 4 / 3. Now let's work on this because the first method gave us a to be equal to the um the the cube root of 81 / 16. Do you think this is the same as that? The answer is yes. Let me show you.
Okay. So this is what the second method gave us. And the first method gave us um something else. So let's get it from here. Now this can be written as 3 / 2 to the^ 4 then everything to the power of 1 / 3 because we are multiplying the two powers. Okay. So we take a step to get a = 3 ^ 4 81 2 ^ 4 16. So now we have 16 to the^ 1 / 3 and that is exactly what we got from the first method. Okay. So the first method gave us um a to be the cube root of 81 over 16. So either of the methods that you use is um very very okay but I believe the second method is faster and easier for math students. Thank you for watching.
Ähnliche Videos
Olympiad Mathematics | Indian | Can You Solve This One?
PhilCoolMath
650 views•2026-06-03
Escaping the Fog
LogicLemurGaming
760 views•2026-06-03
A Brutal Radical Expression Made Easy! The Shortcut Changes Everything.
tamoshop
112 views•2026-06-02
V : jee main /advance class 11 mathematics : Binomial Theorem class-1 ( 29 may 2026 )
dcamclassesiitjeemainsadva9953
125 views•2026-05-29
Is This Pentomino Tileable?
3cycle
241 views•2026-05-30
This Sudoku Has Many Lines!!
CrackingTheCryptic
2K views•2026-05-29
Olympiad Mathematics | Indian Can You Solve This One?
PhilCoolMath
268 views•2026-06-02
Olympiad Mathematics | Indian | Can You Solve This?
PhilCoolMath
669 views•2026-06-02











