CAT 2023 Slot 1 · Quant · Question 3

Name the two roots: let a=\sqrt{5x+9} and b=\sqrt{5x-9}, so a+b=3\left( 2+\sqrt{2}\right)=6+3\sqrt{2}.

Squaring each, {a}^{2}=5x+9 and {b}^{2}=5x-9, so {a}^{2}-{b}^{2}=18.

Since {a}^{2}-{b}^{2}=\left( a+b \right)\left( a-b \right), the difference follows: a-b=\frac{18}{6+3\sqrt{2}}.

Factoring 3 out of the denominator, a-b=\frac{6}{2+\sqrt{2}}.

Rationalising by multiplying top and bottom by 2-\sqrt{2}, a-b=\frac{6\left( 2-\sqrt{2}\right)}{4-2}=3\left( 2-\sqrt{2}\right)=6-3\sqrt{2}.

Adding this to a+b=6+3\sqrt{2}, the surds cancel and 2a=12, so a=6.

Then {a}^{2}=36=5x+9, giving 5x=27.

So 10x=54, and \sqrt{10x+9}=\sqrt{54+9}=\sqrt{63}.

Since 63=9\times 7, this is 3\sqrt{7} — the fourth option.

Shortcut. Multiplying the given sum by its conjugate difference is the whole trick: {a}^{2}-{b}^{2}=18 is a plain number, so the conjugate falls out in one division and the two equations add to give a at once. Note also that 10x+9 is 2\left( 5x+9 \right)-9, so once 5x+9=36 is known the target is 63 directly.