To solve polynomial equations with higher powers, first factor out common terms and apply the zero product rule to break the equation into simpler factors. For the equation x^4 - 27x = 0, factor out x to get x(x^3 - 27) = 0, yielding x = 0 as one solution. Then solve x^3 - 27 = 0 using the difference of cubes identity (a^3 - b^3 = (a-b)(a^2 + ab + b^2)), which gives (x-3)(x^2 + 3x + 9) = 0. The quadratic equation x^2 + 3x + 9 = 0 is solved using the quadratic formula x = [-b ± √(b^2 - 4ac)]/(2a), yielding complex solutions x = (-3 ± 3i√3)/2. The complete solutions are x = 0, x = 3, x = (-3 + 3i√3)/2, and x = (-3 - 3i√3)/2.
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Olympiad Mathematics | Brazilian | Can You Solve This?追加:
Hi, everyone.
Let's provide a complete solution to what we have on the board.
Um, if you look at this, we have the highest power or the higher power of four, right?
And that means that we are to get four solutions.
And it is not every student that will be able to get this right. So, what do we do?
Solution.
We have x to the power of four minus 27 x equals zero.
We are going to carry out factorization first because x is common to the two terms here.
Here we have x to the power of three minus here we're going to get 27 and we equate to zero.
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.
Our zero product rule because we multiply two terms to get zero. So, one of them has to be equal to zero.
Now, we say that x is either zero or x cubed minus 27 equals zero.
When we say x is zero, we have a solution already.
And that is That is our first solution.
So, we need three more solutions.
And that will come from here. x to the power of three minus 27 is equal to zero.
We are expecting three solutions because of the power of three.
So, what do we do?
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.
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.
a cubed - b cubed is definitely a - b multiplied by a squared + ab + b squared.
This is an identity that we must always know.
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.
Then, we open these brackets to multiply this second part.
a squared is x squared - ab is x * 3, which is 3x.
b squared is 3 squared and it is 9.
So, we get 0.
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.
So, we say that x - 3 is either 0 or x squared - 3x.
This is plus, by the way. Got it over there. So, this is plus x squared + 3x + 9 = 0.
So, um from the left-hand side our x is going to be 0 + 3, which is 3.
By the way, we have a solution from here.
So, this is our second solution.
So, we need two more solutions, and the two more solutions will come out of this equation here.
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?
So, I'm going to point this out, then we'll solve it using general formula method.
So, this is our quadratic equation, and we have um A to be 1. A is the coefficient of x squared.
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.
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.
>> [snorts] >> So, that's what we're going to do right away.
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.
This is all over 2 multiplied by 1.
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.
This is all over two.
We go on to get our X equals minus three plus or minus square root of negative 27.
Now, look at one of the mistakes that students will make when solving quadratic equation using formula method.
Look at 9 minus 36. 9 is a perfect square. 36 is a perfect square.
So, students will be tempted to find the square root of 9 separately and the square root of 36 separately.
But, that is wrong.
Yes, that will be wrong.
So, what do we do? We simplify before we find the square root.
This is over two.
Now, look at what I want to do.
I This is X being equal to minus three plus or minus I root 27.
What has happened?
This negative here comes out as our I because square root of negative one is I.
So, this is all over two.
Then, there's a perfect square you know, that we have as a factor of 27 and it's nine, right?
So, we have to simplify that root um 27.
So, our X from here will be equal to minus three plus minus I square root of 9 * 3 all over two.
Interesting, right?
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.
Then, multiply by root three.
And this is all over two.
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.
Now, what do we do? Let's bring the four solutions together, right?
The equation that we have solved is X to the power of four minus 27 X equals zero.
And our first solution is that X is equal to zero.
Okay?
Then, our second solution is that X is also equal to three.
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.
Then, the fourth solution is minus three minus three i root three all over two.
So, these are the four solutions to the equation.
And I believe that you are happy.
And um if you are, then you have to subscribe to my channel, like the video, comment, and share to your friends.
Thank you for always supporting.
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