Olympiad Mathematics | The two solutions obtained | Indian | Can You Solve This One?

[00:00:00]Everyone, do you know how to solve this problem here?

[00:00:08]We want to get the complete okay the real solutions.

[00:00:14]We have 2 square root of a the square root of a and is equal to a. So how do you solve this? You know what? I'm going to solve this in two ways. Just pay attention.

[00:00:28]So the first thing I'm going to do is to square both sides of the equation. So we have the square root of a the square root of a and this is all squared. So this means that the right hand side will also be squared. Now for those of you that would want to ask why do we have to square? Because we have square root on the left. So you have to square to remove the square root which you will do the same to the right even though we do not have square root on the right. So now this is the same thing as 2 to power two * the square root of a square root of a this will also be squared. This is happening from one of the laws of based on one of the laws of indices. Let me explain the law under here.

[00:01:23]The law says that if you have x y and it's raised to the power of let's say b that this can be expressed as x^ b * y to the same power b and this is what I applied to the left hand side. I hope you can understand. So 2^ 2 is 4 and then the square root and that square will go. So that means we we have to multiply this by this. So on the right hand side we have a to the power of two.

[00:01:58]So what do we do again? We are going to have um um square both sides again. But before then multiply these two. We have 4 a right 4 a. Then we have root um a I told you we will multiply this again and then we have we'll square it again. Then the right hand side will be squared for the second time.

[00:02:29]And like I said the square here will work for the three terms. Here we have this this and this. 4 squar is 16. So write your 16. This is 16 right? Let me write this better.

[00:02:44]So we have 4 square to be 16 a² will come down then roo<unk> a² this will take that. So we just multiply this by a and this is equal to a to the power of what? Do you know that the relationship between these two powers is multiplication? So that means we have power 4 over there. Now if you take a step further, you're going to have 16 a cube and is equal to a to the power of 4.

[00:03:16]Next step, bring this one here. We have a to the power of 4 to be equal to 16 a to the power of 3, right? Because the one with the higher power should come down first. So from this point, we want to factoriize. So we're going to bring them together.

[00:03:35]This is -16 a to the^ of 3. Remember there's nothing on the right. So we put zero there.

[00:03:44]And um from the left we have a common factor which is a.

[00:03:49]So that here we have a to the power of three remaining. And here we have 16 a to the power of okay by the way there's still a common factor right? So that means we should do something.

[00:04:04]Okay. Watch me do it. The common factor is a cube, not a to not just a. Okay. A to the^ 3 is common. So here we are going to have 1 a and here we just have 16. Mhm. So this is equal to zero. And we have to apply our zero product rule as we say that either a to power 3 is = 0 or a - 16 is z. So if a to power 3 is 0 it means that a will be equal to the square root of or the cube root of 0 which will still be 0. So a = 0 is a solution already and from this part we have a to be what 0 + 16 and that is 16.

[00:04:54]So there we have another solution. So from the first method we are having these two solutions. But do you think the second method will be faster than this or easier than this? I think the answer is yes. Let's go and look at it.

[00:05:12]Okay. So let's apply the second method here. Second method.

[00:05:20]I believe it's going to be faster. So we have two square root of a square root of a equals a. Remember the first method we had to square both sides twice. Now in this case we understand that if you have square root of x that you can write it as x ^ 1 / 2. Okay. So that is what we know and we are going to use that method to solve it. So this is 2 * a is under one square. So we raise it to the power of 1 / 2. Then this second a is under two square roo. So we raise it again to the power of 1 / 2 * 1 / 2 that will be four [snorts] because it is under two square root sign and we have a on the right. So from here we have two already which is not under any square root sign and then do you know that here we have the same base. We can pick one of them and have 1 / 2 + 1 / 4. Why am I doing that? If you pick one of the bases according to one of the laws of indices, pick one of the bases and add the powers. So, we have two already times we have a already. The LCM of the power is 4. 4 / 2 is 2 * 1 is 2 + 4 / 4 is 1 * 1 is 1. And this is equal to a. You know, we can continue, right? So, if we continue, we're going to have two here, right?

[00:07:02]Multiplying a to the power of 3 / 4. So that this is equal to a and um this dot here means multiplication. So I can actually multiply them to get 2 a to the power of 3 / 4. Bring this to the left. We have minus a. There's nothing on the right anymore. I told you this method will be easier and faster. See? So the next point is that we look for what is common to this two. And it's a. So we're going to bring a out as the common factor. And if a comes out here, we're okay, we're going to have two here first, right? We're going to have two right there. And then we have a to the power of what?

[00:07:50]A to what power? Remember just like we are dividing a to the power of 3 / 4 divided by just a to the power of 1. Oh, sorry I'm writing out of sight. Right?

[00:08:03]Look at it over there. So from there we have the same base. Pick one of them.

[00:08:07]Then we have 3 / 4 - 1. Do you know that? That will give us a to the power of -1 / 4. I hope you know how I arrive at that a to the power of -1 / 4. So that is what we are going to have and from there we will now have here to be raised to the^ 1 / 4 minus this / itself is 1. So we equate to zero and like we always do a is either zero or 2 a to the^ of -1 / 4 is okay -1 is 0. This is already a solution just like we had from the first method. And from this part we can say that 2 a to the power of -1 / 4 is = 1 taking um -1 to the right hand side. So from here we've got to we've got to divide by two divide by two. This will take that out for us. And then we have a to the power of -1 / 4 to be equal to 1 / 2. So what do you think we are going to do again?

[00:09:31]What we going to do is this. We want to um get the value of a. So we have to do away with this negative first. You know what? Let me continue from here.

[00:09:45]Okay. So from here now to remove the negative we're going to have 1 / um how do you call it 1 / a to the^ of 1 / 4 and we still have 1 / 2. Then take the reciprocal of both sides so that we can have a to the^ of 1 / 4 over 1 which is the same thing. And then when you take the reciprocal of 1 /2 you're going to get 2 over 1 which is the same that um the same as two. Then from here to remove the power we raise both sides to the power of four. So four will take that out and a will now be 2 ^ 4 which is 16. And do not forget before now we had a to be zero. So these are the two solutions. And I feel like verifying yes because we have to be sure. Remember the equation is 2 square<unk> of a square root of a = a. If you put zero here, 0 * 2 will definitely give us zero here. So we are saying that we are saying that um a to be 0 satisfies already. If we put 16, we're going to have 2 square root of 16 square root of 16. Will this be equal to 16? want to find out. So the next step we have two multi um square root of 16 * square root of 16 is 4. So we have 16 * 4. Now let's focus on the left.

[00:11:19]This implies we have 2 square<unk> of 64 because 16 * 4 is 64. What is the square root of 64? 8. So here we have 2 * 8 which is 16. Look at it over there. So this also means that a to be equal to 16 satisfies.

[00:11:42]Thank you for watching. If you did not understand the first method, you should understand the second method. See you around.