Amazing Math Olympiad problem| Can you get the three values of x?

JJ ONLINE MATHS CLASS
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A math OLYMPIAD algebra question. The three values of x were obtained. #matholympiadquestions #algebra
AI Generated·May contain errors
Hi, let's quickly get the three values of x. First of all, let's move this to the other side.
If we do that, we have x cubed plus x squared equals 4 divided by 27.
All right, we need a linear equation, so to make this equation linear, let's actually multiply.
through by 27, okay, so if we do that, we are going to have 27 multiplied by x to the power of 3.
plus x squared equals 4, okay; in exponential form, 27 is the same as 3 to the power of 3
multiplied by x to the power of 3 plus x squared is equal to 4 okay let's go ahead and
Open this bracket: if we do, we'll have 3 cubed times x cubed plus 3 cubed times x squared is equal to 4.
All right, now this is the same thing as 3x to the power of 3 plus now this 3 to the power
3 can be written as 3 multiplied by 3 squared, then x squared is equal to 4.
Okay, now at this point we have 3x to the power of 3 and 3x to the power of 3, both of them.
x cubed plus 3 times this same thing as 3x squared is equal to 4
Now we have 3x; we have 3x. Let's assign an alphabet to e, so we say let 3x be equal to y.
and if that is the case, then we will have y to the power of 3
plus 3y squared equals 4; okay, now at this point let's move this 4 to the other side
because what we have here is what we just formed a cubic equation, so we have y to the power of 3
plus 3y squared minus 4 equals 0; now we need two factors of a
Okay, this is a cubic polynomial, so let's use synthetic division. Yeah, let's use synthetic division.
division to obtain the quadratic equation that we will use to multiply by a linear equation
Meanwhile, I'm going to attach a video that explains how to solve your polynomial.
polynomial; you are going to see different methods of solving the polynomial in that
specific video, so I'm going to attach that video here, so after this you can click on it
and watch more on polynomials; at this point, I'm going to use synthetic division to factor it out
the value of our quadratic equation, but before that, let's just try an error and get the first
the root of this equation; now, the first root of this equation—let's look at it; you can see that one
can be used, so if we say that y is equal to one, that means that y to the power of three
which will still give us one plus three times one squared, which is three, still three minus four is
equal to zero because one plus three gives you four and four minus four is zero, so that means
that y is equal to one, that one is a root, so we are going to use this one to divide this
synthetically to obtain the quadratic equation, so if we do that, we are going to have, we have, we are
I'm going to bring out this thing; let's bring out the coefficients of this particular polynomial.
but equations one and three—we're supposed to have y, since it's not there it's zero the
If the coefficient is zero, then the constant is minus four; then we are going to divide this.
Go ahead, okay? So this is how it goes. Let me continue from here because I don't want to
So we have one three zero and minus four, then one is here.
We're going to bring down this one, one times one, we give us one, right? Then you add three plus one.
is four one times four we give you four then you add zero plus four is four one times four we give
you four now minus four plus four we give you zero, so that means that the quadratic equation
What we are looking for is one y squared such that y squared plus four y plus four equals zero.
Therefore, the two factors of this particular equation are what if y is equal to one.
implies that y this plus y crosses to become minus one, okay, so it means that y minus one is
equal to zero, which means that y minus one multiplied by y squared plus four y plus four
is equal to zero, okay. Now we have two values of y here, the first one, I mean, three values, the first
The value is that y minus one is equal to zero, and this has two values: y squared plus four y plus
Four is equal to zero, so for this side we can say that y is equal to what minus one—we cross over.
become plus one, but recall that we are not looking for y, we are looking for x, okay? Now we say three.
x is equal to y, which means that three x is equal to one, so divide by three and divide by three.
so we can say that what our x is equal to one over three, so this is our first value of x, x is
equal to one over three; now for this side of the equation we have y squared plus four y plus four
is equal to zero, so we need two factors of four, but when you add them, you get four again, all right.
and those two factors are two and two, okay, because two plus two will give us four and two
times two will give us plus four, so we can now write this as y squared plus two y plus two y
plus four equals zero; now let's factorize. If we factorize by grouping for here and here,
y is common, y squared divided by y will give us y, two y divided by y will give us two, okay.
Similarly, here what is common is two times y divided by two, which is y, and four divided by two.
Is two equal to zero? Okay, now we can see that what we have outside the bracket is y plus two.
so we have y plus two, all right, multiplied by we have y plus two also in the bracket
y plus two equals zero, so this simply means that we are going to get the same result.
When we expand this, okay, so let's go ahead and do that, but before that, please click on
the subscription button and on your notification bell if this is the first time they are coming
across this channel, okay, so we have y plus two brackets, y plus two, y plus two, multiplied by y
plus two equals zero, so that means y plus two equals zero twice, right, two times
that means y is equal to this minus two, we cross over to become plus two, but we are looking
for x, okay, but x is equal to what? Our 3x is equal to y, so 3x is equal to y. Now that means
We can write that 3x is equal to minus two, so divided by three, divided by three, so we have that.
our x times our x is equal to minus two over three times two; therefore, the values of x are the first
The value of x we obtained is x₁ equals one over three, x₂ equals minus two over
Thank you so much for watching, have a lovely day!
