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

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The discussion focuses on solving the equation \(16a^2b^2 - 48a^2b + 24ab^2 + 100a^2 + 16b^2 - 72ab + 150a - 48b + 100 = 28\) for real number pairs \((a, b)\). Participants engaged in deriving the necessary algebraic manipulations to simplify and solve the equation. The conversation highlights the importance of factoring and using the quadratic formula to find the values of \(a\) and \(b\). The final solutions were not explicitly provided, indicating further exploration is needed.

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Determine the pair(s) of real numbers $(a,\,b)$ that satisfy the equation $16a^2b^2-48a^2b+24ab^2+100a^2+16b^2-72ab+150a-48b+100=28$.
 
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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 $(-3/4,3/2)$.
 
Bravo, Pranav!(Clapping) And thanks for participating!:)
 

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