# Find all real solutions of the given equation.

#### anemone

##### MHB POTW Director
Staff member
Find all real solutions of the equation $$\displaystyle 8x^2-11y^2-40x-8y+4xy+4y^2x+y^4+52=0$$

#### mathmaniac

##### Well-known member
Throw all terms in y to one side and keep the terms in x in the other side.Factorise both and give values for x and get y or the other way.

#### MarkFL

Staff member
Throw all terms in y to one side and keep the terms in x in the other side.Factorise both and give values for x and get y or the other way.
Problems posted in this sub-forum are such that the OP has a solution, and is giving the problem, which they have found interesting or instructional, to the membership here as a challenge to solve.

Responses here should be a well-explained solution (or requests for clarification if the problem statement is vague or misleading).

#### mathmaniac

##### Well-known member
Sorry I am in a hurry to log out.I will make a try on this when I come back...

But surely following what I said leads to the solution,doesn't it?

#### MarkFL

Staff member
Well, the OP is not looking for help, so you really should just wait until you have a full solution to post.

#### anemone

##### MHB POTW Director
Staff member
Find all real solutions of the equation $$\displaystyle 8x^2-11y^2-40x-8y+4xy+4y^2x+y^4+52=0$$
My solution:

First, I group $$\displaystyle 4xy+4y^2x$$ and factor to get:
$$\displaystyle 4xy+4y^2x=4xy(1+y)$$

I notice that $$\displaystyle (1+y)$$ is a factor of $$\displaystyle 4xy+4y^2x$$ and my plan is to rewrite the given expression as the sum of two terms and next equate them to zero.

To do this, I need to group a certain amount of terms and hopefully, there will be the factor of $$\displaystyle (1+y)$$ in them so that I could further simplify the expression.

For $$\displaystyle y^4$$ in the remaining terms of the given expression $$\displaystyle 8x^2-11y^2-40x-8y+\cancel {4xy+4y^2x}+y^4+52=0$$:

I add a $$\displaystyle y^3$$ in the equation and but I will subtract it next to maintain the equality.

$$\displaystyle y^4+y^3=y^3(y+1)$$

For $$\displaystyle -11y^2-8y-y^3$$ in $$\displaystyle 8x^2-11y^2-40x-8y+\cancel {4xy+4y^2x}+\cancel {y^4}+52-y^3=0$$:

I know that if I want to let $$\displaystyle (1+y)$$ be a factor of this cubic equation, I must add a constant of 2 to the equation and therefore I will proceed like this:

$$\displaystyle -11y^2-8y-y^3+2=-\left(y^3+11y^2+8y-2 \right)=-(1+y)\left(y^2+10y-2 \right)$$

Now, for the remaining terms in $$\displaystyle 8x^2-\cancel {11y^2}-40x-\cancel {8y}+\cancel {4xy+4y^2x}+\cancel {y^4}+50-\cancel {y^3}=0$$, I see that I could rewrite it to become:
$$\displaystyle 8x^2-40x+50=2(2x-5)^2$$

Putting all these pieces together yields:
$$\displaystyle 4xy(1+y)+y^3(y+1)-(1+y)\left(y^2+10y-2 \right)+2(2x-5)^2=0$$

$$\displaystyle (1+y)\left(4xy+y^3-\left(y^2+10y-2 \right) \right)+2(2x-5)^2=0$$

$$\displaystyle (1+y)\left(4xy+y^3-y^2-10y+2 \right)+2(2x-5)^2=0$$

Clearly $$\displaystyle x=\dfrac{5}{2}$$ and $$\displaystyle y=-1$$ is one of the valid answers to the problem.

I suspect the given equation can be rewritten as the sum of two terms in different way as well so let's see:

From $$\displaystyle -8y+4xy+4y^2x$$, I simplify it to get:

$$\displaystyle -8y+4xy+4y^2x=4y(-2+x+xy)$$ (*)

We know from a well-known identity that $$\displaystyle xy+x-y-1=(x-1)(y+1)$$

Thus, (*) can be manipulated a bit to obtain the following:
$$\displaystyle -8y+4xy+4y^2x=4y(-1-y+x+xy)+4y^2-4y=4y(x-1)(y+1)+4y^2-4y$$

Now, we know we must rewrite the rest of the given terms with $$\displaystyle (x-1)$$ as one of the factors. If we focus on the terms $$\displaystyle 8x^2-40x$$ from $$\displaystyle 8x^2-11y^2-40x-\cancel {8y+4xy+4y^2x}+y^4+52=0$$, we see that we may rewrite it by having the factor $$\displaystyle (x-1)$$ iff we have a constant of 32 in the equation, like this:
$$\displaystyle 8x^2-40x+32=8(x-1)(x-4)$$

Next, by replacing these two newly rearranged terms within the equation we find that now we have:
$$\displaystyle 4y(x-1)(y+1)+4y^2-4y+8(x-1)(x-4)+y^4-11y^2+32=0$$

or

$$\displaystyle (x-1)(4y(y+1)+8(x-4))+4y^2-4y+y^4-11y^2+32=0$$

$$\displaystyle (x-1)(4y(y+1)+8(x-4))+y^4-7y^2-4y+32=0$$

$$\displaystyle (x-1)(4y(y+1)+8(x-4))+(y-2)^2(y^2+4y+5)=0$$

We can see that $$\displaystyle x=1$$ and $$\displaystyle y=2$$ is another pair of the valid answers to the problem.

I find that there are no other ways to regroup the terms and rewrite it as the sum of two terms and I conclude that the solutions (real) to the problem are

$$\displaystyle (x,y)=\left(\frac{5}{2},-1 \right),\,(1,2)$$.

#### mathmaniac

##### Well-known member
At first sight I thought it was a quadratic equation in two variables...and so there will be infinite solutions...
I blame myself for not reading the question....

Anyway I appreciate your skill of factoring the equation so cleverly.....

And your way of explanation is also cool....

#### Jester

##### Well-known member
MHB Math Helper
The equation is quadratic in $x$ so re-grouping gives

$8\,{x}^{2}+ \left( -40+4\,y+4\,{y}^{2} \right) x-11\,{y}^{2}-8\,y+{y}^ {4}+52 = 0$.

The discriminant when simplified is

$-16(y+1)^2(y-2)^2$

which means that for real $y$, then $x$ will be complex. Since we require the solutions to be real gives the only choices $y=-1$ and $y=2$ as provided.