This presentation masterfully deconstructs complex polynomials by showcasing the versatility of algebraic identities through multiple logical paths. It offers a refined look at the rigorous problem-solving techniques characteristic of high-level competitive mathematics.
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Olympiad Mathematics | Indian | Can You Solve This? | Four solutionsAdded:
Hi.
If you're ready, let's solve this one very quickly.
Okay, we have x to the power of 4 to be equal to 4 to the power of 3 multiplied by 125 x.
So, how do you solve this?
There is x here, there is x over there.
So, there'll be need for us to bring the terms with x together.
But, remember that you you do not divide both sides by two, right? I mean, you don't divide both sides by the x.
So, it's if possible bring x to this side. So, we're going to say x to the power of 4 minus 4 to power 3 multiplied by 125 x is equal to 0.
So, we don't divide by x, we do this so that we can now factor out x.
If x comes out here, we're going to have x to the power of 3.
Then here, we have 4 to the power of 3 multiply multiplied by 125.
The x here is already out.
Okay.
So, we equate to 0.
And now we are multiplying this by this to get 0. So, we can say that it's either x is equal to 0, the one outside, or here x to the power of 3 minus 4 to the power of 3 * 125 is equal to 0.
Okay. So, if we have this, it means that we already have a solution.
This is one of the solutions.
And to get the other solutions, we're going to bring this here, which is x ^ 3 - 4 ^ 3 * 125 and this is equal to zero.
So, from here now, we are expected to have three more solutions.
Yes. So, let's get the three of them.
But, remember that this is x ^ 3 - 4 ^ 3, then multiply by 125 is 5 ^ 3.
So, this is equal to zero.
Now, to go on with this, we have x ^ 3 - 4 * 5.
Right? We can combine this and both of them will have the same power.
Right?
This is equal to zero. So, we go on to get x ^ 3 4 * 5 is 20. And then, we have 20 ^ 3, which is equal to zero.
So, at this point, we have our difference of two cubes.
Right? We have difference of two cubes, and this is an identity in mathematics.
Okay. So, this is difference of two squares.
And um difference of two cubes, rather.
And we know that if we have a cubed - b cubed, this will give us a - b to multiply a squared + ab + b squared.
Right? So, from here now, we're going to have three more solutions.
Remember, we had one solution before now, right?
So, our a minus b now is going to be x minus 20. So, we write x minus 20 then into a squared is going to be x squared then plus um ab our ab is going to be um x times 20 and that is 20x then we have plus b squared and our b squared is going to be 20 squared. So, let's write 20 squared over here.
So, this is equal to this zero.
And then and then to continue from here, we apply zero product rule for the second time.
Okay, and to do that from here we say that x minus 20 is equal to zero or this one is equal to zero.
And from here our x is going to be zero plus 20.
Zero plus 20 is 20.
Okay, so this is our second solution.
The first solution is x equals zero and the second solution is x equals 20. So, I'm going to pick this um expression and equate to zero. Let's do that.
Okay, so from here we have our quadratic equation.
We're going to use quadratic formula to solve it.
A is one, that is the coefficient of x squared.
B is 20, that is the coefficient of x.
And then c is 20 squared.
And what is the formula? The formula is x equals minus b plus or minus square root of b squared minus 4 ac all over 2 a.
And from here now, our x is going to be in place of minus b, we'll put minus 20 plus or minus b squared is going to be 20 squared.
Right?
And then we have minus 4 * 1 * c. Our c is 20 squared. So, we put 20 squared for the c.
And everything is over 2 * 1, which is the same as 2.
Now, to continue with this, we have our x to be minus 20 plus or minus 20 squared here and 20 squared there.
So, we write 20 squared as a common factor.
Right? Then we open bracket and have 20 squared divided by itself is 1 minus this 20 squared is already out, so we have 4 * 1 and that is 4.
This is all over 2.
To go on, we are going to have x to be equal to minus 20 plus or minus we have 20 squared multiplied by negative three because 1 minus 4 is negative three.
And this is all over two.
Now, from here, we can split what we have there under the root so [snorts] that x will be minus 20 plus or minus we have the square root of 20 squared multiplied by the square root of negative three.
Okay, so you can always split this like this.
And this is now over two.
So to go ahead with this, we are going to have x to be equal to Okay, uh x will be equal to we have 20 plus or minus By the way, let me remove this. The square root and the square will go. So here we have 20 then multiply by square root of three multiply by i. This negative is going to be i because square root of negative one is i.
And this is all over two.
But then we can arrange this in a better way.
So we have x to be 20. This is minus 20.
Minus 20 plus or minus We have 20 times i, that is 20 i. Then we have root three.
And this is all over two.
So let's us um simplify this better so that our x will be equal to two into minus 20 is minus 10 plus or minus two into 20, that's going to be 10 i. Then we have root three.
But mind you, this is a two-in-one kind of solution.
Now let's bring the four solutions together.
Okay, so from our calculation, the four answers the four solutions to this equation are x equals zero the first solution x equals 20 the second solution then we have X equals minus 10 plus 10 I root three the third solution then we have X to be equal to minus 10 minus 10 I then we have root three the fourth solution so these are the four solutions to the given equation thank you for watching if you're new here consider subscribing to my channel like comment and share
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