IPMAT Indore 2024 · Quantitative Ability · Question 39

Step 1 — substitute. Note {x}^{2}={\left| x \right|}^{2} for every real x, so put t=\left| x \right|\ge 0:

{t}^{2}-10t-56=0

Step 2 — factorise. Looking for two numbers with product -56 and sum -10: these are -14 and +4.

\left( t-14 \right)\left( t+4 \right)=0 \Rightarrow t=14 \text{or}t=-4

Step 3 — reject the impossible root. t=\left| x \right|\ge 0, so t=-4 gives nothing.

Step 4 — unfold the modulus.

\left| x \right|=14 \Rightarrow x=14 \text{or}x=-14

Check: 196-140-56=0 ✔ for both.

Answer: Option 3 — 2

Shortcut. Each positive root of the t-quadratic gives two x; each non-positive root gives none (and t=0 would give exactly one).

So the count is decided purely by the signs of the roots, which you can read off without factorising: product =-56<0 means one positive and one negative root, hence 2\times 1=2 solutions. Three seconds, no factorisation at all.

Option (b) 4 is the trap for anyone who doubles both roots; option (d) 1 for anyone who forgets the negative x.