Brazilian | Olympiad Mathematics | Can you solve this?

Added: 2026-05-05

[00:00:00]Hi everyone.

[00:00:03]If you're ready, let's solve this.

[00:00:08]Y to the^ 3 over 2 = 4.

[00:00:16]Okay, we are expected to get three solutions from here. So what do we do?

[00:00:23]Cross multiply, right? Because this is over one. Now y cub to the power um y cub * 1 is the same y cub.

[00:00:34]Then we have 2 * by 4 and that is 8.

[00:00:40]So by now you can see it's possible for us to get three solutions.

[00:00:45]The first thing we will do is we write y ^ 3 to be equal to 2 to the^ 3.

[00:00:53]Now do not equate the bases right because we are getting the three solutions. So write y cub - 2 cub now equal to zero.

[00:01:08]If they ask you what has he done, what would you say? What I just did is to move 2 to the^ 3 to the left hand side and in place of 2 to the^ 3 only zero remains there and 2 ^ 3 becomes negative on the left hand side. So from here we have difference of two cubes and we know that if we have a cube minus b cub that this can be written as a - b * a 2 + a b + b 2. This is what we call an identity in mathematics.

[00:01:58]So what do we do from here? In place of a minus b, we are going to write y - 2.

[00:02:07]Meaning that our a is y and our b is 2.

[00:02:12]And then we open the brackets to write this a 2 is y 2 + a b that's going to be y * 2 and is 2 y. Then + b 2 is going to be y 2.

[00:02:29]No, B² is going to be 2^ 2 and 2^ 2 is 4. So this is equal to 0.

[00:02:37]And at this point you should know what to do, right? We are going to multiply this and this to get zero. So we should apply what we call zero product zero product rule.

[00:02:56]So um when do we apply this rule? We apply this rule when we have two terms to multiply.

[00:03:04]Let's say you have um PQ to be equal to zero.

[00:03:10]Remember PQ is P * Q, right? For this statement to be true, either of these should be zero or even both of them can be zero because 0 * 0 will give 0. 0 * Q will give 0. and then p * 0 we still give zero. So either of them is 0. The same thing we're going to apply over here.

[00:03:35]Okay. So we say that y - 2 is either 0 or y^ 2 + 2 y + 4 is = z.

[00:03:50]Now let's work on the left hand side first. So from here our y is = 0 + 2 and um okay so from here our y is = 2. This is one of the solutions.

[00:04:09]Now look at that that quadratic equation I have to bring it down here. We have y^ 2 + 2 y + 4 = 0. This this is a quadratic equation and um we have y.

[00:04:29]Okay. So we have um the coefficient of y as our a. We're going to use quadratic formula for this. And the formula has a b c. The coefficient of y^ 2 is 1. That is our a. The coefficient of y is our b and is equal to 2 from here. Then the constant C is equal to 4. So what do we do?

[00:04:57]Get our quadratic general formula and it's equal to y = - b + or minus the square root of b^ 2 - 4 a c all over 2 * a.

[00:05:14]And you have to know that the y here is not constant, right? It depends on the unknown variable in the equation that you have in this equation. The unknown is y. So that's why y is the subject of the formula. If the unknown variable is x, then here should have been x. So let's put in our a, b, and c. So that y will now be in place of minus b. I'll put -2 + or minus the square roo<unk> of b^ 2 is 2^ 2 - 4 * 1 * 4 a is 1 c is 4 so this is this is over 2 * 1 so we're going to continue with this and I believe that um you're getting what we are doing so our y now is equal to we Write - 2 + or minus and there 2^ 2 is 4 - 4 * 1 * 4 is 16. The whole of this is over two.

[00:06:25]We have to do something right. Subtract what we have here. So y will be -2 + or minus. Then we have the square root of -2 because 4 - um 16 is -2 and this is all / 2.

[00:06:45]What I'm going to do is to separate the negative from here to pick out the negative from here. So y will be -2 plus or minus the square roo<unk> of 12 * the<unk> of -1 as we divide all through by two.

[00:07:04]Right? Now you look at the 12 over there. We can write 12.

[00:07:11]We can write 12 as 4 * 3. Then multiply by i. the square root of -1 is i and this is all over 2.

[00:07:26]So this means that y = - 2 plus or minus roo<unk> of 4 is 2 * this i. Right? So that we have 2 i. Then we are here to write roo<unk>3. We're writing roo<unk>3 because 3 is not a perfect square and everything is over two.

[00:07:47]Now guess what I'm going to do? Y = 2 into - 2 is -1 + or - 2 into 2 I. That will give us I <unk>3.

[00:07:59]And this is a two in one solution. Let's bring the three solutions together. If you remember before now we got one solution and it's that it's um y = 2. So let's call that y1.

[00:08:15]So we're going to get our y2 which is um from here -1 plus or minus we have i oh this is going to be plus alone <unk>3 then our y3 the third solution y3 is = -1 by the way okay so we have -1 - i <un three. So these are the three solutions to the equation. Thank you for watching.

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