Nice Olympiad Mathematics | The complete solution.

Added: 2026-05-09

[00:00:01]Let me show you how to solve this problem here.

[00:00:06]We have 3 x to the power of 4 equals 2 to the power of 4 plus 2 to the power of 5.

[00:00:17]We are going to solve this problem completely.

[00:00:21]Okay, so what do you do?

[00:00:24]Let's work on the right-hand side.

[00:00:27]3 x to the power of 4 to be equal to from here 2 to the power of 4 is a common factor.

[00:00:39]Right?

[00:00:40]So, I'll bring 2 to the power of 4 out as a common factor.

[00:00:45]2 to the power of 4 divided by itself is 1 plus 2 to the power of 5 divided by 2 to the power of 4, that will be equal to 2.

[00:00:58]Right?

[00:00:59]Okay, I believe you're getting what I'm doing now. So, this is 3 x to the power of 4 and is equal to 2 to the power of 4 right? 1 + 3 1 + 2 is 3.

[00:01:13]So, at this point, we can easily reduce our equation as we divide both sides by 3.

[00:01:24]3 will cancel itself and cancel itself.

[00:01:27]So, we have x to the power of 4.

[00:01:31]x to the power of 4 is equal to 2 to the power of 4.

[00:01:37]But then because we have the power of 4, we are expected to have four solutions.

[00:01:44]So, how then do we go about that?

[00:01:46]Let's break what we have here.

[00:01:49]You know, this is the same thing as x squared squared, right?

[00:01:55]And then, we can do the same thing to this and get 2 squared.

[00:02:00]Then we have square 2.

[00:02:02]This is because we are expected to multiply the powers from one of the laws of indices that says that a to the power of m to the power of n is the same thing as a to the power of mn because you are to multiply the powers, right?

[00:02:21]Okay, let me remove this.

[00:02:26]Okay, so to go on with this, we can bring this to the left.

[00:02:32]Yes, we can do that so that we have x squared squared minus By the way, 2 squared is going to be 4, so let me use 4 here.

[00:02:43]Okay, minus 4 squared is equal to 0.

[00:02:47]And this is our popular difference of two squares.

[00:02:51]Right? Because a squared minus b squared is always a plus b times a minus b.

[00:03:01]Okay, the one that comes first does not really matter. It can be a minus b and then times a plus b. It does not um matter.

[00:03:11]So, our a plus b now is going to be x squared plus b is 4.

[00:03:20]Into a minus b, that's going to be x squared and we have minus 4.

[00:03:26]So, this is all equal to 0.

[00:03:31]This is all equal to 0. And at this point, we are going to apply our zero product rule.

[00:03:38]Zero product zero product rule because we are multiplying these two to get zero. So, either of the factors here must be equal to zero. Okay? So, to break it into two we will have either x squared plus four is equal to zero or x squared minus four is equal to zero.

[00:04:08]Okay?

[00:04:10]Um let's pick it up from here.

[00:04:14]Okay, so let's work on this first part.

[00:04:20]And uh we have x squared plus four. We want to remove four by so we minus four and it's equal to zero minus four as well.

[00:04:31]Okay? When I get there, I will do the same.

[00:04:35]Now, from here we have x squared.

[00:04:38]Four minus four is gone, so we have this to be equal to minus four.

[00:04:44]And we want to solve it to get um two solutions from here because we know that this is a quadratic um equation.

[00:04:53]So, to get our value of x from here we have to take the square of the square root of both sides.

[00:05:01]Then we have plus or minus the square root of minus four.

[00:05:07]So that from here this one can just take this out and our x is equal to plus or minus the square root of minus four.

[00:05:19]But then we can do something here. Let's separate what we have here so that x will be plus or minus square root of four times the square root of negative one.

[00:05:31]Root four times root negative one will give root negative four.

[00:05:36]Yes. So, that means we are still in line. And luckily for um for us four is a perfect square, so we can get the square root of four.

[00:05:47]So, our x is equal to square root of four is two multiplied by square root of negative one is imaginary. So, we write I.

[00:05:57]But, this is plus or minus, right? So, to continue with this, we have our X to be equal to 2i or -2i.

[00:06:10]So, from here we have two solutions.

[00:06:13]Although these are not real solutions, they are complex roots, right?

[00:06:18]So, remember we left out one of the factors.

[00:06:23]Okay, we left out one of the factors and it is X to the power of two minus four equals zero. So, from here again we are expected to have two more solutions.

[00:06:36]So, we have X squared minus four. Now, because it's minus four, to remove the four, we add another four.

[00:06:45]Then we have zero plus the four we are adding.

[00:06:49]Now, -4 + 4 is gone. So, we just have X to the power of two and it's equal to zero plus four. That will give us four.

[00:07:00]Now, let's use difference of two squares here.

[00:07:03]Yes, let's use difference of two squares. This is X squared to be equal to two squared because four is the same as two squared, right?

[00:07:12]Okay. Now, to continue with this, we have to bring this to the left and we have X squared minus two squared to be equal to zero.

[00:07:24]This is our difference of two squares and from the identity for difference of two squares, this is X minus two into X plus two to be equal to zero.

[00:07:39]Now, at this point, we can apply zero product again.

[00:07:43]So, it is x - 2 to be zero or x + 2 to be zero.

[00:07:52]So, from here now, our x is going to be zero plus two and that is two.

[00:07:57]Or from here, x will be zero minus two and that is minus two.

[00:08:03]So, from here we have two more solutions. So, what we will do now is to bring the four solutions together.

[00:08:12]Yes, let's bring the four solutions together. Remember, we got x >> [clears throat] >> We got x to be equal to 2i before. Let's call this our x1.

[00:08:27]Then, our x2 is -2i, right?

[00:08:34]-2i and then our x3 is our two from here. These twos are x3.

[00:08:44]And then, the fourth one is -2. So, these four are the solutions to the given equation.

[00:08:53]Thank you for watching.

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