## Saturday, April 11, 2015

### 16x²y²-48x²y+24xy²+100x²+16y²-72xy+150x-48y+100=28

Determine the pair(s) of real numbers $(a,\,b)$ that satisfy the equation $16x^2y^2-48x^2y+24xy^2+100x^2+16y^2-72xy+150x-48y+100=28$.

At first glance,  this seems like we need to use the modular arithmetic method to solve for $(x,\,y)$ but wait a minute! We need to find not the integers but real for $(x,\,y)$, so nope, modular arithmetic is out of the question...

Okay, now, what plan do we have? We got to have a plan in order to start, and we are not going to stare at the ceiling in order to gain the idea, we need to look at the given equation again and see if there is something we could do about it...

$16x^2y^2-48x^2y+24xy^2+100x^2+16y^2-72xy+150x-48y+100=28$

Note that we could rewrite the LHS of the equation by treating and grouping them as quadratic expressions in $y$ and $x$:

Please note also that we could regroup the terms that have the same coefficients, such as the pairs of

[MATH]\color{yellow}\bbox[5px,purple]{16x^2y^2}[/MATH] and [MATH]\color{yellow}\bbox[5px,purple]{16y^2}[/MATH];

[MATH]\color{yellow}\bbox[5px,red]{48x^2y}[/MATH] and [MATH]\color{yellow}\bbox[5px,red]{48y}[/MATH];

[MATH]\color{yellow}\bbox[5px,orange]{100x^2}[/MATH] and [MATH]\color{yellow}\bbox[5px,orange]{100}[/MATH]

so that

$(16x^2y^2+16y^2)-(48x^2y+48y)+(100x^2+100)+24xy^2-72xy+150x=28$

$16y^2(x^2+1)-48y(x^2+1)+100(x^2+1)+24xy^2-72xy+150x=28$

You might want to ask, does this approach work? Oh...we wouldn't know unless we try...at least this is something that we observed and we can simplify the LHS of the equation, let's see where it takes us:

$(x^2+1)(16y^2-48y+100)+24xy^2-72xy+150x=28$

$4(x^2+1)(4y^2-12y+25)+6x(4y^2-12y+25)=28$

$(4y^2-12y+25)(4(x^2+1)+6x)=28$

See it now? We can still "simplify" the LHS of the expression by completing the square of the two quadratic expressions:

$4\left(y^2-3y+\dfrac{25}{4}\right) \cdot 4\left(x^2+\dfrac{6x}{4}+1\right)=28$

$\left(y^2-3y+\dfrac{25}{4}\right)\left(x^2+\dfrac{3x}{2}+1\right)=\dfrac{28}{16}$

$\left(\left(y-\dfrac{3}{2}\right)^2+\dfrac{25}{4}-\dfrac{9}{4}\right) \left(\left(x+\dfrac{3}{4}\right)^2+1-\dfrac{9}{16}\right)=\dfrac{28}{16}$

$\left(\left(y-\dfrac{3}{2}\right)^2+4\right) \left(\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{16}\right)=\dfrac{28}{16}$

YES! Observe that $4\cdot \dfrac{7}{16}=\dfrac{28}{16}$, so, the only solution set for $(x,\,y)$ is $\left(-\dfrac{3}{4},\,\dfrac{3}{2}\right)$.