CAT 2024 Slot 1 · Quant · Question 2

Squaring, {\left( a+b\sqrt{n}\right)}^{2}={a}^{2}+n{b}^{2}+2ab\sqrt{n}, and this must equal 29-12\sqrt{5}.

Matching the rational and irrational parts, {a}^{2}+n{b}^{2}=29 and 2ab\sqrt{n}=-12\sqrt{5}.

The natural reading is n=5 and ab=-6, and then {a}^{2}+5{b}^{2}=29 gives a=-3, b=2.

That value is -3+2\sqrt{5}\approx 1.47, which is positive as required.

But the question asks for the maximum of a+b+n, and this form is not the only way to write the same number.

Since 2\sqrt{5}=\sqrt{20}, the same square root is also -3+1\cdot \sqrt{20}, with a=-3, b=1 and n=20.

Checking, {a}^{2}+n{b}^{2}=9+20=29 and 2ab\sqrt{n}=-6\sqrt{20}=-12\sqrt{5}, so this is valid.

That gives a+b+n=-3+1+20=18, and no larger n is possible because b must stay an integer.

Shortcut. Pulling a factor inside the radical — b\sqrt{n}=\sqrt{{b}^{2}n} — trades a smaller b for a much larger n. Since n carries far more weight in the sum, always push the coefficient inside before totalling.