Question

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

Answer 1

Let $A=4x^2+6x+4$

$\Rightarrow A.(4y^2-12y+25)=28$

$\Rightarrow 4y^2-12y+25-\frac{28}{A}=0$

Since there is a unique pair $(x,y)$ the discriminant $b^2-4ac$ of the quadratic formula must be equal to zero.

$\Rightarrow {12}^2-4\times 4\times (25-\frac{28}{A})=0$

$\Rightarrow 144-400+\frac{448}{A}=0$

$\therefore A=\frac{7}{4}$ on simplification

So, $A=4x^2+6x+4=\frac{7}{4}$

$\Rightarrow 4x^2+6x+4-\frac{7}{4}=0$

$\Rightarrow 4x^2+6x+\frac{9}{4}=0$

Using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-6\pm \sqrt{6^2-4\times 4\times \frac{9}{4}}}{2\times 4}=\frac{-3}{4}$ since again the discriminant is zero.

Substituting $A$ into the original equation, we get

$\frac{7}{4}(4y^2-12y+25)=28$

$\Rightarrow 4y^2-12y+9=0$

$\Rightarrow {\left(2y\right)}^2-2\times 2y\times 3+{\left(3\right)}^2$

$\Rightarrow {(2y-3)}^2=0$

$\therefore y=\frac{3}{2}$

$\therefore$ the unique pair of real numbers $(x,y)$ that satisfy the equation

$(4x^2+6x+4)(4y^2-12y+25)=28$ is $(\frac{-3}{4},\frac{3}{2})$