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Find all real solutions of the given equation.

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
Find all real solutions of the equation \(\displaystyle 8x^2-11y^2-40x-8y+4xy+4y^2x+y^4+52=0\)
 

mathmaniac

Active member
Mar 4, 2013
188
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Active member
Mar 4, 2013
188
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

Administrator
Staff member
Feb 24, 2012
13,775
Well, the OP is not looking for help, so you really should just wait until you have a full solution to post.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,676
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

Active member
Mar 4, 2013
188
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
Jan 26, 2012
183
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.