Given two different quadratic equations in x and y separately, find an ordered pair which satisfies an eqation containing their products.

Question

Let $P(x) = 4(x)^2 +6x+4$ and $Q(x)= 4(y)^2 - 12y +25$ Find the unique ordered pair of real numbers $(x,y)$ that satisfies $P(x).Q(x) = 28$

Please tell me how to answer such questions

Answer 1

I suppose that the numbers are real, otherwise the problem has infinite solutions.

Let's begin with $$4 x^2+6 x+4=h;\;4 y^2-12 y+25=\frac{28}{h}$$ The two equations satisfy $$(4 x^2+6 x+4)(4 y^2-12 y+25)=h\cdot \frac{28}{h}=28$$ The first equation has two solutions $$x_1=\frac{1}{4} \left(-3-\sqrt{4 h-7}\right),\;x_2=\frac{1}{4} \left(-3+\sqrt{4 h-7}\right)$$ They are coincident if $4h-7=0;\;h=\frac{7}{4}$

So we get $x=-\frac{3}{4}$

The other equation, in $y$, becomes $$4 y^2-12 y+25=\frac{28}{\frac{7}{4}}\rightarrow 4 y^2-12 y+9=0$$ which has again only one solution $y=\frac{3}{2}$

The pair which solves the problem is $\left(x=-\dfrac{3}{4};\;y=\dfrac{3}{2}\right)$

Remark. It was a key information to know that the solution is unique.

Hope it helps