Olympiad Mathematics | Indian | Can you solve this?

Añadido: 2026-05-11

[00:00:00]Hi everyone.

[00:00:03]If you are ready, let's provide the solution to this problem here.

[00:00:08]Solution.

[00:00:13]Okay.

[00:00:15]We have x to the power of 3 divided by 3 to be equal to the square root of 81.

[00:00:25]Okay, so let's break it down the way it should.

[00:00:30]We have x to the power of 3 multiplied by 1 because we know this is over 1.

[00:00:35]So x to the power of 3 * 1 is x to the power of 3 and it's equal to 3 multiplied by the square root of 81.

[00:00:46]And 81 is a perfect square, so we can find its square root.

[00:00:51]So x to the power of 3 is equal to 3 multiplied by the square root of 81, which is 9.

[00:01:00]So at this point, what do we do?

[00:01:03]We have to write x to the power of 3 to be equal to 3 multiplied by 3 squared.

[00:01:10]Because 9 is 3 squared, right?

[00:01:14]Now our x to the power of 3 is equal to 3 to the power of 3 because there's an invisible one over there.

[00:01:23]And if you pick one of the bases, you're going to add the powers.

[00:01:28]Now from here we are not after um getting only solution.

[00:01:34]We want to get three solutions from here.

[00:01:38]So why are we getting three solutions?

[00:01:40]Because of the power 3.

[00:01:43]Okay, the power of the variable determines the number of solutions you're going to have.

[00:01:48]Now to get three solutions we will have to bring this to the left.

[00:01:55]So we have x to the power 3 - 3 ^ 3 to be equal to 0.

[00:02:03]Because what is on the right has been shifted to the left.

[00:02:08]And at this point, we have difference of two cubes. This is an identity, right?

[00:02:14]If we have m cubed minus n cubed and this is given to you, the identity is just m minus n multiplied by m squared plus mn plus n squared.

[00:02:35]Okay, so this is the identity.

[00:02:40]And we're going to apply this, use this to get our value of x or the values of x.

[00:02:47]Now, let's continue with that. Remember that our in place of this x now we wrote m, in place of three, we wrote n.

[00:02:56]So, our m minus n becomes x minus three.

[00:03:03]Right? Then into m squared is going to be Oh, the m squared is going to be three x squared, right?

[00:03:12]Then we have plus mn, which is going to be x multiplied by three.

[00:03:19]Then we have plus n squared and it's three squared. So, the whole of this is equal to zero.

[00:03:28]And here we have x minus three one of the factors. Then we have x squared plus three x plus nine. Three squared is nine and x times three is three x. So, this is equated to zero.

[00:03:48]And at the end of the day, we can easily apply our zero product rule because we are multiplying these two to get zero, right?

[00:03:59]And every time we multiply two terms together zero, one of the terms must be equal to zero.

[00:04:07]Okay, so in this case we'll say that it's either x minus three equals zero.

[00:04:14]Right? Either x minus three equals zero or x squared plus three x plus nine equals zero.

[00:04:27]Now, from the left-hand side we have a linear equation.

[00:04:31]And on the right part we have um a quadratic equation. So, linear equation will give a solution and quadratic equation will give two solutions. From this part our x will just be zero plus three.

[00:04:48]As simple as it is. And um this means that x is equal to three. So, this is our first solution.

[00:04:59]Okay, I said first because we're expecting two more. And the two solutions we are expecting will come from this quadratic equation.

[00:05:09]Okay, so let's pick it and then solve it.

[00:05:14]Okay, so from here we have our quadratic equation.

[00:05:19]And it has a one b three and c to be nine.

[00:05:29]Okay? Okay, those of you that do not know how I got the a b c a is the coefficient of x squared and it's plus one.

[00:05:37]B is the coefficient of x and it's plus three.

[00:05:41]C is a constant and it's plus nine.

[00:05:44]That's positive nine.

[00:05:46]>> [snorts] >> Now, the next thing is to bring down our quadratic formula.

[00:05:51]It is x equals minus b plus minus we have b squared minus four ac.

[00:06:01]So, everything is over two times a.

[00:06:05]So, the next point is to substitute into this um interesting formula and we have our x to be minus b turns to minus three plus or minus b squared is going to come down here b squared minus [snorts] four times um four times one cuz a is one then multiplied by nine.

[00:06:33]c is nine and this is all over two times one.

[00:06:38]So, to continue we simplify what we have under the the root. So, that's x will be minus three plus or minus we have the square root of three squared nine then four times nine that is 36.

[00:06:56]So, this is all over two times one.

[00:06:59]And two times one is still going to give us two.

[00:07:04]Right? So, we are still going to simplify this by subtracting.

[00:07:12]x will now be minus three plus or minus the square root of nine minus 36 that will give us minus Okay, that will give us minus um 27.

[00:07:27]Okay, so this will be over two and then x will now be minus three plus or minus square root of 27 multiplied by square root of negative one.

[00:07:43]What I've done is to pick out the negative from the 27.

[00:07:49]So that we can process root 27.

[00:07:53]Although 27 is not a perfect square, so we need to bring out the perfect square in there, and that is 9 * 3.

[00:08:02]>> [snorts] >> Then the square root of -1 is imaginary, so we put I.

[00:08:08]We are dividing this by 2.

[00:08:11]So that our X from here will be -3 plus or minus the square root of 9 here is 3. Multiply this 3 by I, we have 3i. Then we are still having root 3 over there.

[00:08:27]And everything here is over 2.

[00:08:31]Yeah. So as a matter of fact, we have a two-in-one solution.

[00:08:36]So what we'll do now is to bring out the four solutions. I mean, the three solutions.

[00:08:44]But remember that this will produce two.

[00:08:47]One is going to be positive here, and the other one is going to be negative.

[00:08:52]So we got X to be equal to 3 as our first solution.

[00:08:57]Then the second solution, right? Let me write it here. Our X2 is from here we have -3 + 3i root 3.

[00:09:12]Right? So this is all over 2, our second solution.

[00:09:17]And the third, X3, is -3 -3i root 3.

[00:09:26]Everything over 2.

[00:09:28]So these are the two, I mean, the three solutions to the equation.

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