MHB Solve for a & b: Real Number Pairs $(a,\,b)$

[anemone]

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

 

[Saitama]

Spoiler

Rewrite the equation as:
$$16b^2(a^2+1)-48b(a^2+1)+100(a^2+1)+24ab^2-72ab+150a=28$$
$$\Rightarrow 4(a^2+1)(4b^2-12b+25)+6a(4b^2-12b+25)=28$$
$$\Rightarrow (2a^2+3a+2)(4b^2-12b+25)=14$$

The minimum value of $2a^2+3a+2$ is $7/8$ at $a=-3/4$ and that of $4b^2-12b+25$ is $16$ at $b=3/2$. Clearly, the pair $(a,b)$ is the $(-3/4,3/2)$.

 

[anemone]

Bravo, Pranav!(Clapping) And thanks for participating!:)