JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
The sum of modulus of all non-real roots of the equation $(x^{2}+4x+6)(2x^{2}+8x+14)=12$ is equal to $\_\_\_\_\_\_$ .
Quadratic Equation · Equations Reducible to Quadratic
Answer$6$
Solution
Concept. The second factor is $2(y+1)$ with $y=x^2+4x+6$, so the equation is a quadratic in $y$. Formulas used. (1) $|p+iq|=\sqrt{p^2+q^2}$Step 1 - Quadratic in $y$. $2y(y+1)=12$, i.e. $y^2+y-6=(y+3)(y-2)=0$. Step 2 - $y=2$. $(x+2)^2=0$: real roots only. Step 3 - $y=-3$. $(x+2)^2=-5$: $x=-2\pm i\sqrt5$. By (1), each has modulus $\sqrt{4+5}=3$. Step 4 - Sum. $3+3=6$. Answer: 6