CAT 2023 Slot 2 · Quant · Question 2

Label the three powers A={2}^{4{x}^{2}}, B={2}^{2{x}^{2}+x+16} and C={2}^{2x+30}, so the equation reads A-B+C=0.

Look at the product of the outer two: AC={2}^{4{x}^{2}+2x+30}.

Its square root is \sqrt{AC}={2}^{2{x}^{2}+x+15}.

Comparing exponents, B={2}^{2{x}^{2}+x+16}=2\times{2}^{2{x}^{2}+x+15}=2\sqrt{AC}.

So the equation is A+C=2\sqrt{AC}.

Rearranging, A-2\sqrt{AC}+C=0, which is exactly {\left( \sqrt{A}-\sqrt{C}\right)}^{2}=0.

Therefore A=C, so the exponents must be equal: 4{x}^{2}=2x+30.

Rearranging, 4{x}^{2}-2x-30=0, and halving, 2{x}^{2}-x-15=0.

The discriminant is 1+120=121, so x=\frac{1\pm 11}{4}, giving x=3 or x=-\frac{5}{2}.

Their sum is 3-\frac{5}{2}=\frac{1}{2} — the third option.

Shortcut. Only the sum of the roots is asked, so solving is unnecessary: once the equation becomes 2{x}^{2}-x-15=0, Vieta gives the sum as \frac{1}{2} directly. The middle exponent being the average of the outer two, plus one, is the signal that a perfect square is hiding.