CAT 2025 Slot 3 QA Question 13: If $f(x) = (x^2 + 3x)(x^2 + 3x + 2)$, then the sum of all real roots of the equation $\sqrt{f(x)+1} = 9701$, is

Question

If $f(x) = (x^2 + 3x)(x^2 + 3x + 2)$, then the sum of all real roots of the equation $\sqrt{f(x)+1} = 9701$, is

Accepted Answer

Let $t = x^2 + 3x$. Then
$f(x) = t(t + 2)$
$f(x) + 1 = t^{2} + 2t + 1 = (t + 1)^{2}$
$\sqrt{f(x) + 1} = 9701$
$|t + 1| = 9701$
So, $t+1 = 9701$ or $t+1 = -9701$.
$t = 9700\ 	ext{gives}\ x^2 + 3x - 9700 = 0$
$	ext{Sum of roots} = -3$
$t = -9702\ 	ext{gives}\ x^2 + 3x + 9702 = 0$
$	ext{Its discrim

If f(x)=(x2+3x)(x2+3x+2)f(x) = (x^2 + 3x)(x^2 + 3x + 2)f(x)=(x2+3x)(x2+3x+2), then the sum of all real roots of the equation f(x)+1=9701\sqrt{f(x)+1} = 9701f(x)+1​=9701, is

If $f(x) = (x^2 + 3x)(x^2 + 3x + 2)$ , then the sum of all real roots of the equation $\sqrt{f(x)+1} = 9701$ , is