This video demonstrates how to solve the equation x^4/x + x^3/x = 36 by applying exponent rules (a^m ÷ a^n = a^(m-n)) to simplify it to x^3 + x^2 = 36, then factoring using the difference of cubes and squares identities to find three solutions: x = 3, x = -2 + 2i√2, and x = -2 - 2i√2, while noting that x = 0 must be rejected as it would cause division by zero in the original equation.
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Olympiad Mathematics | The three solutions | Can You Solve This?Added:
Hi.
How many solutions do you think we can have from here?
We have x to the power of 4, right?
Then plus x to the power of 3 over x equals 36.
Okay, this is very simple.
We can split what we have here, right?
So, if we split this, we're going to have x to the power of 4 over x plus x to the power of 3 over x. This is equal to 36.
Okay. Then the next thing we're going to do is to pick one of the bases, right? From this very law, a to the power of m divided by a to the power of n is a to the power of m minus n.
Pick one of the bases and you do what?
You subtract the powers because it's division here, and it's the same division over there. So, we have x to the power of 4 minus 1 plus we have x to the power of 3 minus 1 because there's an invisible power of 1 here, the same over there.
So, the whole of this now is equal to 36.
Meaning that we have x to the power of 3 plus x to the power of 2 be equal to 36.
Now, from here, we have to solve it now.
It means that we can only have three solutions and not four.
Okay.
Um let's work it.
We look at 36.
Can we express 36 in the base I mean to be in this form?
The answer is yes.
Because from 36 we can have 27.
27 is 3 to the power of 3, so it can be like this.
Then what do we add 36 to get what we add to 27 to get 36?
It is 9, right? So 9 can be in this form. So we are settled.
We have x to the power of 3 plus there we have x to the power of 2 to be equal to 3 is um 27 is 3 to power 3 plus here we have 3 to the power of 2.
Now bring this to the left. We have x to the power of 3 minus 3 to the power of 3.
Right? This is together as a group.
Then we have plus we're going to get another group for x squared.
Remember x is positive, x squared is positive, right? But this comes here to become negative 3 squared.
And on the right-hand side we have zero.
From here now we have difference of two cubes and difference of two squares.
Remember that x a cube minus b cube is equal to a minus b multiplied by a squared plus ab plus b squared.
Okay, so this is the identity for the difference of two squares.
Then the difference of two cubes rather and you know that for the difference of two squares.
So let's use this to express what we have there now.
Our x cubed - 3 cubed will now be x - 3.
Okay, starting from here, a is x, b is 3. Multiplying x squared, which is going to be here for the a squared, we have x squared now.
Then ab, that's going to be x * 3 and it's 3x.
Plus, here we have what? b squared.
And the b squared is going to be 3 squared, which will give 9.
This is for the difference of two cubes.
Then plus, we go over to the difference of two squares.
And that is going to be x - 3 into x + 3.
This is our popular difference of two squares.
And everything here is equal to zero because of this.
So, from here now, we have a common factor, which we're going to bring out.
And it is x - 3.
Yes, we have x - 3 here and there, right?
So, the next is this one plus this.
This one plus this will now be in this bracket because x - 3 is already out.
So, we write our x squared plus 3x plus 9 then plus x plus 3.
So, this is what we have and we equate to zero.
Okay, so from here we have our x - 3.
Then in here, we have x squared plus x, 3x + x is 4x.
Then 9 plus 9 plus 3 is 12.
This is equal to zero.
Now we apply our zero product rule.
So that X minus 3 is zero or X squared plus 4 X plus 12 is equal to zero.
From the left-hand side, our X will be equal to 3.
This is our first solution.
Now we need two more solutions to this equation, right?
And what are we going to get?
If you're going to solve this, you have to know your A, which is 1, know your B, which is 4, and know your C, which is 12.
Because in the equation we're going to use, we have A B C.
So, not equation, in the formula that we're going to use, right?
What is the formula?
X equals minus B plus or minus we have B squared minus 4 AC over 2 A.
This is what we call the quadratic general formula.
So we put in our ABC.
So that X will be equal to we have minus B, which will now be minus 4 plus minus B squared that's going to be 4 squared.
Right? Then minus 4 times 1 for A, then times 12 for the C.
So all of this is over 2 multiplied by 1.
So if we go on, we have X to be minus 4 plus minus 4 squared is 16 minus 4 times 12 is 48.
Remember this is all over two.
So, we have to continue as we subtract 16.
Or we take 48 out of 16. So, it will give us negative value.
Our X is minus four plus or minus square root of negative 32. That is 16 minus 48.
And we divide by two. This is two.
Okay, so we go on.
We're going to get X to be what? Minus four plus or minus square root of 32.
Oh, I didn't put the negative, right?
So, I have to multiply it by the negative one.
Then we divide this by two.
Break your 32 because 16 is a factor of 32.
We have to work on that as we have um minus four plus or minus the square root of 16 times two, then multiply by the square root of negative one, which is imaginary.
Remember that we're still dividing this by two.
So, we go on. Our X is going to be minus four plus or minus square root of 16 is four.
Then multiply by this I, we're having four I. Then multiply by root two.
Because this two is still under the square root sign. It is not um a perfect square.
And this is divided by two.
So, we have a two in one solution, but we can simplify it.
So, that our X will be equal to two into minus four is minus two plus or minus two into four I, that will be two I as we multiply by two.
This is a two in one kind of solution because of the plus or minus. Now, let's bring the three solutions together.
Okay, so these are the Um okay, this is the original equation, right?
And now, these are the solutions. X equals three.
Our first solution.
Then, we have X two to be equal to minus two plus two I root two.
This is our second solution. So, the next one I want to have negative here.
So, we have X three, which should be minus two minus two I.
Then, we have root two.
So, these are the three solutions to this equation.
But, is there a way I will solve this equation here and I'll have X to be zero?
Okay, X to be zero.
Ooh.
We added to this, right? But, it will not be accepted because there's no way the value of X here can be zero. So, even if you solve it and you get this value, you will have to reject it. So, these are the three values of X.
Thank you for watching.
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