Olympiad Mathematics | Brazilian | Can You Solve This?

Ajouté : 2026-05-15

[00:00:00]Hi, everyone.

[00:00:03]Let's provide a complete solution to what we have on the board.

[00:00:09]Um, if you look at this, we have the highest power or the higher power of four, right?

[00:00:19]And that means that we are to get four solutions.

[00:00:24]And it is not every student that will be able to get this right. So, what do we do?

[00:00:29]Solution.

[00:00:32]We have x to the power of four minus 27 x equals zero.

[00:00:38]We are going to carry out factorization first because x is common to the two terms here.

[00:00:45]Here we have x to the power of three minus here we're going to get 27 and we equate to zero.

[00:00:53]If you open the bracket, you are going to have um You're going to get this right back. So, what do we do from here? We carry out our zero products.

[00:01:10]Our zero product rule because we multiply two terms to get zero. So, one of them has to be equal to zero.

[00:01:20]Now, we say that x is either zero or x cubed minus 27 equals zero.

[00:01:30]When we say x is zero, we have a solution already.

[00:01:34]And that is That is our first solution.

[00:01:37]So, we need three more solutions.

[00:01:40]And that will come from here. x to the power of three minus 27 is equal to zero.

[00:01:47]We are expecting three solutions because of the power of three.

[00:01:52]So, what do we do?

[00:01:54]We realize that three to the power of 3 is 27. So, let's write x to power 3 - 3 to power 3 to be equal to 0.

[00:02:07]So, that the next point we will, you know, the next step we're going to take is to apply our zero our difference of two cubes identity.

[00:02:17]a cubed - b cubed is definitely a - b multiplied by a squared + ab + b squared.

[00:02:32]This is an identity that we must always know.

[00:02:37]So, now we, in place of a - b x - 3 will come in because a is representing x b is for 3. So, a - b is x - 3.

[00:02:52]Then, we open these brackets to multiply this second part.

[00:02:57]a squared is x squared - ab is x * 3, which is 3x.

[00:03:05]b squared is 3 squared and it is 9.

[00:03:09]So, we get 0.

[00:03:11]Now, we are multiplying both of them again to get 0, right? So, for the second time, we are going to apply our zero product rule again.

[00:03:24]So, we say that x - 3 is either 0 or x squared - 3x.

[00:03:35]This is plus, by the way. Got it over there. So, this is plus x squared + 3x + 9 = 0.

[00:03:45]So, um from the left-hand side our x is going to be 0 + 3, which is 3.

[00:03:53]By the way, we have a solution from here.

[00:03:57]So, this is our second solution.

[00:04:00]So, we need two more solutions, and the two more solutions will come out of this equation here.

[00:04:07]This is a quadratic equation because of the power of two, and because the power is two, I mean the highest power is two, we're going to multiply No, we're going to get two solutions because of the power of two, right?

[00:04:22]So, I'm going to point this out, then we'll solve it using general formula method.

[00:04:31]So, this is our quadratic equation, and we have um A to be 1. A is the coefficient of x squared.

[00:04:42]B is 3, and our C is 9. Okay? So, what we'll do now is to write the formula, which is X equals minus B plus minus B squared minus 4 AC divided by 2 * A.

[00:05:04]So, once you're able to remember the formula, what you will do is to just put in the value or the values of A, B, and C.

[00:05:15]>> [snorts] >> So, that's what we're going to do right away.

[00:05:17]And our X becomes minus 3 plus minus B squared as 3 squared minus 4 * 1 * 9 because A is 1, and C is 9.

[00:05:33]This is all over 2 multiplied by 1.

[00:05:37]And we process this further to get X being equal to minus 3 plus or minus 3 squared is 9, and 4 * 9 is 36.

[00:05:49]This is all over two.

[00:05:52]We go on to get our X equals minus three plus or minus square root of negative 27.

[00:06:00]Now, look at one of the mistakes that students will make when solving quadratic equation using formula method.

[00:06:11]Look at 9 minus 36. 9 is a perfect square. 36 is a perfect square.

[00:06:16]So, students will be tempted to find the square root of 9 separately and the square root of 36 separately.

[00:06:24]But, that is wrong.

[00:06:26]Yes, that will be wrong.

[00:06:28]So, what do we do? We simplify before we find the square root.

[00:06:32]This is over two.

[00:06:34]Now, look at what I want to do.

[00:06:37]I This is X being equal to minus three plus or minus I root 27.

[00:06:43]What has happened?

[00:06:45]This negative here comes out as our I because square root of negative one is I.

[00:06:54]So, this is all over two.

[00:06:56]Then, there's a perfect square you know, that we have as a factor of 27 and it's nine, right?

[00:07:05]So, we have to simplify that root um 27.

[00:07:10]So, our X from here will be equal to minus three plus minus I square root of 9 * 3 all over two.

[00:07:23]Interesting, right?

[00:07:26]So, X now is minus three plus or minus square root of three is three. Multi- square root of nine is three multiplied by I, we have three I.

[00:07:36]Then, multiply by root three.

[00:07:39]And this is all over two.

[00:07:41]So, from here we have um two in one solution because of the plus or minus, one is going to be positive, the other negative.

[00:07:50]Now, what do we do? Let's bring the four solutions together, right?

[00:07:58]The equation that we have solved is X to the power of four minus 27 X equals zero.

[00:08:07]And our first solution is that X is equal to zero.

[00:08:12]Okay?

[00:08:14]Then, our second solution is that X is also equal to three.

[00:08:23]And our fourth our third solution is what we just got and it's equal to minus three plus three i root three all over two. This our second and our third solution.

[00:08:42]Then, the fourth solution is minus three minus three i root three all over two.

[00:08:52]So, these are the four solutions to the equation.

[00:08:57]And I believe that you are happy.

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