IPMAT Indore 2026 · Quantitative Ability · Question 25

Step 1 — multiply the two quadratics out.

\left({x}^{2}+px+2 \right)\left({x}^{2}+qx-2 \right)

={x}^{4}+\left( p+q \right){x}^{3}+\left( pq-2+2 \right){x}^{2}+\left( 2q-2p \right)x-4

={x}^{4}+\left( p+q \right){x}^{3}+pq\,{x}^{2}+2\left( q-p \right)x-4

Note the {x}^{2} coefficient is simply pq — the +2 and -2 cancel.

Step 2 — match coefficients.

p+q=a \ldots \left( i \right)

pq=3a-4 \ldots \left( ii \right)

q-p=a-1 \ldots \left( iii \right)

Step 3 — solve (i) and (iii) for p and q. Adding and subtracting:

q=\frac{2a-1}{2}, p=\frac{1}{2}

Step 4 — substitute into (ii).

\frac{1}{2}\cdot \frac{2a-1}{2}=3a-4 \Rightarrow 2a-1=12a-16 \Rightarrow 10a=15 \Rightarrow a=\frac{3}{2}

Exactly one value works.

Answer: Option 4 — 1

Shortcut. Use the two linear equations in p,q first — they pin both unknowns in terms of a with no quadratics at all. Only then impose the product condition, which becomes a single linear equation in a. Going via pq first drags you into unnecessary algebra.