IPMAT Indore 2022 · Quantitative Ability · Question 35

Step 1 — split on the critical points x=-4 and x=4.

Case A: \left| x \right|\ge 4. Then \left| x+4 \right|+\left| x-4 \right|=2\left| x \right|, so the equation becomes

{x}^{2}+2\left| x \right|-35=0

Put t=\left| x \right|\ge 4: {t}^{2}+2t-35=0\Rightarrow \left( t+7 \right)\left( t-5 \right)=0\Rightarrow t=5 (rejecting t=-7).

Since 5\ge 4 ✔, this gives x=\pm 5.

Case B: -4<x<4. Then \left| x+4 \right|=x+4 and \left| x-4 \right|=4-x, which sum to 8. The equation becomes

{x}^{2}+8-35=0 \Rightarrow{x}^{2}=27 \Rightarrow x=\pm 3\sqrt{3}\approx \pm 5.196

But \left| \pm 5.196 \right|>4, which contradicts the assumption \left| x \right|<4. So Case B yields no roots.

Step 2 — collect the roots and square them. The roots are x=5 and x=-5:

{5}^{2}+{\left( -5 \right)}^{2}=25+25=50

Answer: Option 3 — 50

Shortcut. Learn the identity for this shape:

\left| x+a \right|+\left| x-a \right|=\left\{\begin{matrix}2\left| x \right|, & \left| x \right|\ge a \\ 2a, & \left| x \right|\le a\end{matrix}\right.

It converts the equation into two clean quadratics with no sign bookkeeping.

The essential discipline is checking each case against its own interval. Case B produces x=\pm 3\sqrt{3}\approx \pm 5.2, which lies outside \left( -4,4 \right) and must be discarded — and it is exactly those discarded values that generate the wrong option (a) 74=25+49 and (d) 148.

Answer 25+25=50.