Find all real solutions of the given equation.

[anemone]

Find all real solutions of the equation \(\displaystyle 8x^2-11y^2-40x-8y+4xy+4y^2x+y^4+52=0\)

 

[mathmaniac]

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]

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]

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]

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

 

[anemone]

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]

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]

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.