Olympiad Mathematics | The complete solution | Can you solve this?

[00:00:00]Okay.

[00:00:01]Can you solve this problem here on your own without any help?

[00:00:08]Can you?

[00:00:09]We have the square root of x plus two to the power of three equals 64 to the power of 1/2.

[00:00:20]Now, when I say solve it, I mean you should solve it completely.

[00:00:24]You provide the complete solution to the problem.

[00:00:30]Okay, so the first step we're going to take is that if you have the square root of a, you can write it as a to the power of 1/2, right?

[00:00:40]Good. So, because you can do that we are going to write this as x plus two.

[00:00:48]This is raised to the power of 1/2.

[00:00:51]Remember that there's still power of three outside here.

[00:00:57]So, everything here is equal to 64 to the power of 1/2.

[00:01:02]And from the law that I explained earlier this is the same thing as the square root of 64 because of the power of 1/2.

[00:01:12]And luckily for us, 64 is a perfect square.

[00:01:16]Okay, so because 64 is a perfect square we are going to do what we are going to do. So, here we have x plus two to the power What if I write three here?

[00:01:30]And I take 1/2 outside.

[00:01:35]Right? And I take 1/2 outside. Okay, let me write the 64 again. This is 64 to the power of 1/2.

[00:01:44]Right? This and this can go. Yes.

[00:01:49]Okay, I think that's the best thing we should do.

[00:01:52]>> [snorts] >> So, on the left now we have x plus three x plus two rather, to the power of three and it's equal to 64.

[00:02:03]And what is 64 for us?

[00:02:05]64 is also 4 to the power of three. So, here we can now write our 4 to power 3 as we have from the same power on both sides.

[00:02:18]Now, if you equate the bases, you're correct, but not completely correct, since the question says we should produce the complete solution. So, we are going to bring 4 to power 3 to the other side. x + 2 to the power of three minus 4 to the power of three is equal to zero.

[00:02:42]And from here, right? We can now apply our um identity, which is a cubed minus b cubed being equal to a a plus a minus b first, and then it multiplies a squared plus ab plus b squared.

[00:03:06]Right? So, this is an identity.

[00:03:09]>> [snorts] >> And if we are going to go by this identity now, our a is x + 2.

[00:03:16]Right? Our a is x + 2 and our b the value of b So, our a is x + 2 and b is 4. So, in place of a now, we're going to put x.

[00:03:28]Okay, this is x + 2.

[00:03:31]Close that.

[00:03:33]Right? Okay, we don't have to close it.

[00:03:35]Then, minus b will now become minus 4.

[00:03:40]So, we close it. Then, we open.

[00:03:43]a squared is x + 2.

[00:03:47]This is to the power of two, squared.

[00:03:50]Then, plus a is still x + 2.

[00:03:55]And let's multiply B which is four.

[00:03:58]Then we have plus B squared. B squared will now be what? Four squared.

[00:04:03]Four squared is equal to 16. So we write 16 here.

[00:04:08]We close this and we equate to zero.

[00:04:12]Catch your zero over there.

[00:04:14]And I believe you know the expansion here. So this is X plus two minus four and it's equal to X minus two.

[00:04:24]Right? So let's expand this. The expansion of this will give us um we have X squared plus um four X plus four.

[00:04:39]That is the expansion of X plus two all squared.

[00:04:43]Then here we open the bracket to have plus X times four is four X.

[00:04:50]Then two times four, that is eight.

[00:04:53]And then we have plus 16.

[00:04:56]I hope we can still see what I'm writing.

[00:04:59]So we have plus 16 and everything is equal to zero.

[00:05:04]Now let's simplify what we have.

[00:05:07]X minus two is still a common factor.

[00:05:12]It's still a factor by the way.

[00:05:14]>> [snorts] >> Then we have X squared there.

[00:05:17]Four X plus four X will give us eight X.

[00:05:22]Then four plus um 16, that is 20. 20 plus eight, that will be 28. So here we have 28.

[00:05:32]And there's nothing else, so we equate this to zero.

[00:05:39]Here now we will apply our zero product rule.

[00:05:43]So it's either X minus two gives us zero.

[00:05:48]Or what we have there gives us zero.

[00:05:52]But here, let's say that x is equal to zero plus two, and that will give two.

[00:05:58]So from here, we have our first solution.

[00:06:02]To get the other solutions, let's bring this one down.

[00:06:06]x squared plus eight x plus 28 equals zero.

[00:06:14]And we'll use our quadratic formula to solve it.

[00:06:19]>> [snorts] >> Okay, so for the formula, we are going to have a which is equal to one.

[00:06:27]B is eight. A is a coefficient of x squared. B is a coefficient of x.

[00:06:33]And c is 28, the constant.

[00:06:38]So the formula is x equals minus b plus minus b squared minus four ac all over two times a.

[00:06:51]Once you're able to recall the formula, what you do is direct substitution.

[00:06:57]So our x will now be minus eight plus or minus the square root of b squared. You know, it's going to be eight squared. Then minus four times one times 28.

[00:07:15]Okay.

[00:07:16]A is one and c is 28. All of this is over two multiplied by one.

[00:07:25]Now, let us um proceed. X will be minus eight plus or minus the square root of >> [snorts] >> eight squared is 64.

[00:07:37]Then four times 28. Four times eight is 32.

[00:07:41]Then four times eight. Four times two is 8 + 3 that is 11.

[00:07:48]112 and we divide by 2.

[00:07:53]Let's continue.

[00:07:56]Okay, so from here, this is X then equal to -8 plus or minus the square root of -48.

[00:08:05]64 - 112 is 148 -148.

[00:08:10]This is over 12 over 2.

[00:08:14]So, our X will now be -8 plus or minus.

[00:08:21]Okay, so we have the square root of 48.

[00:08:25]Oh, you can see that I didn't put negative. So, we multiply this by the square root of -1.

[00:08:31]Everything is over 2.

[00:08:35]Okay, so this is X equals -8 plus or minus the square root of 48 is the same as square root of 16 * 3.

[00:08:46]16 * 3 will give us 48 and we multiply by >> [snorts] >> square root of -1 is I.

[00:08:53]So, all of this is over 2. So, X will now be um we have -8 plus or minus square root of 16 is 4 then multiply by root 3 multiply by I.

[00:09:09]Everything is over 2.

[00:09:11]So, that if we move on, we're going to have X to be -8 plus or minus square root of um Okay, 4 * I that is 4 I. We have root 3.

[00:09:25]Then, divide all through by 2.

[00:09:28]Now, 2 can divide the numerators.

[00:09:31]So, X is 2 into -8 is -4.

[00:09:37]2 into 4 i that will be 2 i and then we multiply by root 3.

[00:09:43]So this is a two-in-one solution.

[00:09:47]To bring the three solutions together, we got x before to be two.

[00:09:53]Then we have x again from here minus 4 plus 2 i root 3.

[00:10:02]Right? Then the third solution x 3 will now be That will be minus 4 minus 2 i and we have root 3. So these are the three solutions to the equation given.

[00:10:19]Thank you for watching.