Sharp Olympiad Mathematics | How I Solved it in two ways.

[00:00:01]Hi, everyone.

[00:00:04]Can you solve this um equation here?

[00:00:09]2a over the square root of 2a equals 3.

[00:00:16]So, what do you think you're going to do first?

[00:00:20]Cross multiply, right?

[00:00:22]So, that's if you multiply this by that, you're going to have um 3 root 2a.

[00:00:31]Then we believe that this is over 1, so 2a * 1 is going to be 2a.

[00:00:37]So, what do I do again?

[00:00:40]We have to remove the square root sign, so we have 3 root 2a on the left.

[00:00:47]To remove the square root sign, we just have to square it.

[00:00:51]And then here we have 2a which we'll also square.

[00:00:58]Now, the next thing is that we apply this.

[00:01:03]If you have mn to the power of 2, you know, this is the same thing as m squared * n squared.

[00:01:15]So, here now we're going to have 3 squared * the square root of 2a squared.

[00:01:22]And everything is 2a * 2a. That is the expansion.

[00:01:29]Now, from the left-hand side, 3 squared is 9.

[00:01:34]Then the square root and the square can cancel out.

[00:01:37]So, we just have 2a on the left multiplying 9.

[00:01:42]And here we have 4a squared.

[00:01:46]And I'm thinking, what again should we do?

[00:01:49]Multiply what we have on the left. 9 * 2a, that's going to be 18a, which is equal to 4a squared.

[00:01:58]Then the next step is that we write 4 a squared first.

[00:02:03]Okay, write 4 a squared. Then this one here comes here to become -18 a.

[00:02:11]Okay, this is me trying to bring everything to the same side. So that means on the other side we have zero.

[00:02:20]Now, what should we do?

[00:02:22]If you want, you can reduce this equation by dividing both sides or dividing up through by two.

[00:02:30]We have two here, so we have 2 a squared.

[00:02:34]Then minus two into 18 is nine, so we have 9 a.

[00:02:38]Zero divided by two is zero.

[00:02:42]So what do you think is common now?

[00:02:45]Um a is common, right? So we bring out a.

[00:02:51]Here we're going to have 2 a minus here nine remains.

[00:02:56]This is equal to zero. Once you have factorized and you're not sure you're correct, try to expand it in your mind.

[00:03:04]a multiplied by 2 a, that is 2 a squared.

[00:03:07]a multiplied by nine, that is 9 a.

[00:03:10]So this means that we are still in line.

[00:03:13]The next point is that since we are multiplying the two terms to get zero, it's either this is zero or this is zero.

[00:03:23]So if I pick a to be zero, I already have a solution.

[00:03:29]Right? But then the second part is 2 a minus nine equals zero.

[00:03:37]So that what we will do from here is to collect terms.

[00:03:43]2 a will be equal to zero plus nine, and that is nine.

[00:03:48]Now, to get the value of a, we divide this by two and divide the other side by two.

[00:03:56]And um from here, we have A on the left, which is going to be equal to 9 over 2.

[00:04:03]So, our A is 9 over 2.

[00:04:08]We have two solutions from the first method. We're going to solve this in two ways.

[00:04:13]The first method, we got A to be 0 and we got A to be 9 over 2. Now, let's apply the second method. I believe it's going to be faster and easier for you.

[00:04:29]Okay, so let's apply our second method.

[00:04:34]So, um what do we do from the second method?

[00:04:39]We're going to have to take the square of both sides.

[00:04:46]So, we have um 2A over root 2A.

[00:04:51]We'll take the square immediately.

[00:04:54]Then we have 3 squared.

[00:04:56]Why are we taking the square almost immediately?

[00:05:00]Because we want to remove this um square root here.

[00:05:04]But then you should understand that if you have I think I've explained this before.

[00:05:09]X over Y squared you can split this to get X squared over Y squared.

[00:05:20]Okay, this is very, very possible.

[00:05:23]So, applying this to the left, I'm going to have 2A squared over the square root of 2A and this is also squared.

[00:05:35]So, that on the right we have nine.

[00:05:39]Now, from here, what do you do?

[00:05:41]The expansion of this will give you um 4A squared.

[00:05:49]Then this one will remove this. So, we divide by 2 a.

[00:05:55]So, at the end of the day, we are going to have nine on the right.

[00:06:00]Interesting, right? I believe this one is faster and uh you know, easier than the first method.

[00:06:08]2 into 4 is 2.

[00:06:11]a a into a squared that will give us 1 a. So, that this is equal to nine.

[00:06:18]Okay? And then, to get your value of a, we have to divide both sides by two.

[00:06:24]This will take this out for us.

[00:06:27]And then, we have a that is equal to 9 over 2.

[00:06:33]Do you remember what you got from what we got from the first method?

[00:06:37]We got a to be zero or um, 9 over 2. So, the first method gave two solutions, and the second method gave us one solution.

[00:06:50]Now, let's put in these values and see if they will satisfy.

[00:06:57]>> [snorts] >> Okay, so.

[00:06:59]Remember, we got um, a to be a to be zero or 9 over 2. Now, if you put zero, you're going to have 2 * 0 over the square root of 2 * 0.

[00:07:15]This will be what? Zero over the square root of zero, which is zero over zero.

[00:07:23]And zero over zero will not give us three.

[00:07:26]Since we have been three on the other side.

[00:07:29]So, it means that our a to be equal to zero is not a solution.

[00:07:36]Now, let's put put the second one, which is 9 over 2.

[00:07:41]2 a now will now be 2 * 9 over 2.

[00:07:46]Then divided by the square root of We have 2 A, so that is going to be 2 again * 9 over 2.

[00:07:57]So, what will this give us? Will it give three?

[00:08:00]Let's confirm it.

[00:08:02]Two can remove this, so we have nine divided by um two can go into this as well, and we are going to have the square root of nine.

[00:08:13]Right?

[00:08:15]So, this means that we have nine.

[00:08:20]Divided by the square root of nine is three.

[00:08:24]So, nine over three is three.

[00:08:26]So, this means that we are very correct to say that X Okay, we are solving in terms of A.

[00:08:33]To say that A equals um nine over two satisfies the equation.

[00:08:41]Thank you for watching, and see you in the next video.

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