<s>[INST] Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$ Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.
Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$

I can solve this question graphically.
$$P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28$$
$$4x^2+6x+4=\frac{28}{4y^2-12y+25}$$
I drew the graph of $4x^2+6x+4$ and found it is a upward parabola with minimum value at $(-\frac{3}{4},\frac{7}{4})$ and I drew the graph of $\frac{28}{4x^2-12x+25}$. I found that it is a bell shaped curve whose maximum value occurs at $(\frac{1}{4},\frac{7}{4})$. So I found that the common abscissa is $y=\frac{7}{4}$ which occurs at $x=-\frac{3}{4}$ for the first curve and at $x=\frac{1}{4}$ for the second curve, so solution should be $(-\tfrac{3}{4},\tfrac{1}{4})$
But I do not  know how to solve it algebraically.
 [/INST] Define function $f:\Bbb R^2\to R$ as following
$$f(x,y)=P(x)Q(y)-28=(4x^2+6x+4)(4y^2-12y+25)-28$$
It is continuous and differentiable as a polynomial. Since solution $f(x,y)=0$ is unique it means that point $(x,y)$ must be an extremum of this function.
Find partial derivatives and $\delta$:
$$f'_x=(8x+6)(4y^2-12y+25)$$
$$f''_{xx}=8(4y^2-12y+25)$$
$$f'_y=(4x^2+6x+4)(8y-12)$$
$$f''_{yy}=8(4x^2+6x+4)$$
$$f''_{xy}=(8x+6)(8y-12)$$
$$\delta(x,y)=(f''_{xy})^2-f''_{xx}f''_{yy}$$
And solve $$\begin{cases}f'_x=0\\f'_y=0\end{cases}$$
$$\begin{cases}(x+\tfrac68)(4y^2-12y+25)=0\\
(4x^2+6x+4)(y-\tfrac32)=0\end{cases}$$
$$\begin{cases}
x=-\frac34\\y=\frac32
\end{cases}$$
We have $\delta(x,y)<0$ and $f(-\tfrac34,\tfrac32)=0$ so it is an extremum and unique solution to $f(x,y)=0$.
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