Olympiad Mathematics | Indian | Can You Solve This One?

[00:00:00]Hi everyone.

[00:00:04]If you're ready, let's go through this one. Here we have the square root of 2 y to the power of 3 = 1 / 8.

[00:00:17]Okay, how do we solve this problem here?

[00:00:22]Um, the first thing we're going to do is to work on this arrangement here. Do you know that the square root of a is the same thing as a ^ 1 / 2? I hope you will know about that right. So let me remove this.

[00:00:41]So if you know about that it means that um okay we can express this as 2 y raised to power 1 / 2 and then everything is still to the power of 3 and is equal to 1 /8.

[00:01:04]Do you agree with this?

[00:01:07]Okay. If you do, we also know that the powers here the the the relationship between the two powers is multiplication. So because of that the one that comes first does not really matter. So we can write um okay can write this as 2 y right 2 y into in bracket raised to the^ of 3 then I am going to take this 1 / 2 outside and then here I have 1 / 8 now to remove that power from there 1 /2 it means that everything that I have here will be raised to the power of two. So that the whole of this again will be raised to the power 1 / two. So this and this are going to go. So on the left hand side see what we have now. We have 2 y all to the power of 3 to be = 1 / 64.

[00:02:16]1 2 is 1 and 8 2 is what? 64.

[00:02:22]So what again do you think we can do from here? You look at what we have on this side. So this is 2 y to the power of um 3 being equal to 1 to the^ of 3 / 4 ^ 3 because 4 to power 3 will give us 8. And from one of the laws of indices, we can combine these two because they have the same powers. So 2 y to the power of 3 will be equal to 1 / 4 to the power of what? 3.

[00:03:03]Okay. Now what again should we do? Let's bring this to the left hand side so that we can have difference of two cubes. So 2 y ^ 3 - 1 / 4 ^ 3 is = 0 cuz I have moved what we have on the right to the left. So zero remains there.

[00:03:29][snorts] So looking at this now we having um difference of two cubes.

[00:03:35]Imagine that you have a - bub a - b * a 2 + a b + b 2. Okay. So this is the identity here and um our a is going to be 2 y. So here we have 2 y - b which is 1 / over 4.

[00:04:05]Then in the second we're going to have 2 y right we're going to have 2 y raised to power 2 + a is still 2 y that's going to multiply b which is 1 / 4 then we have + b² which is 1 / 4 to the power of 2.

[00:04:30]Close this right and everything will be equal to zero.

[00:04:38]Now let's try to simplify. We have 2 y - 1 / 4 into bracket. We remove the square root.

[00:04:49]Now we remove this um bracket within this one. Now 2 y^ 2 is 4 y^ 2 + um 2 will go into 4 1* no 2 will go into four two times. So here we have y / 2 then plus 1 / 4^ 2 is 1 / 16.

[00:05:13]So we equate to zero.

[00:05:17]Now from [snorts] here we apply our product rule. Since we are multiplying two terms to get zero. So we say that either this is zero or everything here is equal to 0.

[00:05:32]So here now we have um 2 y - 1 / 4 is = 0.

[00:05:40]So that 2 y is = 1 / 4.

[00:05:46]And um to get the value of of y we have to multiply both sides by 1 / 2. 1 / 2 * 2 y is = 1 / 4 * 1 / 2. So that two will remove this um two and then we have y to be = 1 /8. This is our first solution.

[00:06:14]So we're going to go back to the top and then we work on the second factor there.

[00:06:21]So we are back here and we are going to work on this. We have 4 y^ 2 + y / 2 + um 1 / 16. We equate it to zero. So we have to know what the LCM is. The LCM is 16. So 16 * 4 that's going to be 64 y^ 2 + now 16 * everything here is going to be 8 y because this two will divide 16 to give 8 then multiply by y 8 y then when you multiply 16 by everything here you're going to have one as we equate to zero.

[00:07:06]So from this point we have quadratic equation and we're going to solve it using our quadratic formula.

[00:07:15]So a is 64, b is 8 and c is 1. So let's bring down the equation that we are going to use which is y = - b + - we have b² - 4 a c this is all over 2 a and then from here y is going to be - b plus or minus okay we substitute already so in place of minus b we put min -8 8 plus or minus. We have um here we have b ^ 2 which is going to be 8 2 and 8 2 is 64 minus we have 4 * a our a is 64 right okay 64 * 4 what would that give us here we have 16 take 1 we have um 24 that would be 256 right so here we have 256 6.

[00:08:26]Okay. And everything is still over 2 * by what? 64. And 2 * 64 is 128. So let's write 128 here.

[00:08:40][snorts] Okay. So y now will be - 8 plus or minus<unk> of -192 / 128.

[00:08:54]So to break it down we have - 8 plus or minus<unk> 192 * the square root of -1.

[00:09:05]Okay, I brought out the negative and we're dividing by 128.

[00:09:10]So y will now be - 8 plus or minus the square root of 192 is 64 * 3. Then square root of -1 is i / 128.

[00:09:28]So if we break this down further, we're going to have y to be - 8 + minus the square root of 64 is 8.

[00:09:39]Multiply by i. We have 8 i. Then multiply by this roo<unk>3. We have roo<unk>3. Everything is over 128.

[00:09:50]Right?

[00:09:52]Okay. So from here we can um reduce this to get y = um 8 into - 8 is -1 + or minus here we're going to have 8. Okay I <unk>3 8 into 128 is 16. So this is a two in one solution.

[00:10:17]So to bring the three solutions down here, we will now have um y to be = 1 / 8 as the first solution.

[00:10:29]Then our second solution is what -1 + i<unk> 3 / 16. This is coming from the last part. Then y3 is -1 - i <unk>3 / 16. So these are the three solutions to the equation.