Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$

Question

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 $P(x)Q(y)=28$

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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.

Answer 1

Note that we have $$P(x)=4x^2+6x+4=4\left(x+\frac 34\right)^2+\frac 74\ge \frac 74$$ and $$Q(y)=4y^2-12y+25=4\left(y-\frac 32\right)^2+16\ge16$$

Answer 2

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 $f(x,y)=0$.