CAT 2025 Slot 3 · Quant · Question 11

Both brackets are built from the same expression, so put u={x}^{2}+3x.

Then f\left( x \right)=u\left( u+2 \right)={u}^{2}+2u.

Adding 1 completes a square: f\left( x \right)+1={u}^{2}+2u+1={\left( u+1 \right)}^{2}.

So \sqrt{f\left( x \right)+1}=\left| u+1 \right|, and the equation reads \left| u+1 \right|=9701.

Since u+1={x}^{2}+3x+1, this means {x}^{2}+3x+1=9701 or {x}^{2}+3x+1=-9701.

The first gives {x}^{2}+3x-9700=0, whose discriminant 9+38800=38809 is positive, so it has two real roots.

By Vieta those roots sum to -3. (They are 97 and -100, since \sqrt{38809}=197.)

The second gives {x}^{2}+3x+9702=0, whose discriminant 9-38808 is negative, so it has no real roots and contributes nothing.

The sum of all real roots is therefore -3.

Shortcut. Whenever a product of two expressions differing by 2 appears, adding 1 makes a perfect square. After that the sum of roots comes from Vieta alone — the value 9701 never has to be used, only its sign.