IPMAT Indore 2025 · Quantitative Ability · Question 23

Step 1 — name the roots. Let them be \alpha and p\alpha.

Step 2 — write Vieta’s formulas for 8{x}^{2}-2kx+k=0:

\text{Sum:}\alpha \left( 1+p \right)=\frac{2k}{8}=\frac{k}{4}\text{Product:}p{\alpha}^{2}=\frac{k}{8}

Step 3 — eliminate \alpha. From the sum, \alpha =\frac{k}{4\left( 1+p \right)}. Put that into the product:

p\cdot \frac{{k}^{2}}{16{\left( 1+p \right)}^{2}}=\frac{k}{8}

Cancel one k (it is non-zero) and cross-multiply:

\frac{pk}{16{\left( 1+p \right)}^{2}}=\frac{1}{8}\Rightarrow pk=2{\left( 1+p \right)}^{2}\Rightarrow k=\frac{2{\left( 1+p \right)}^{2}}{p}

Step 4 — match the option form. Notice that

{\left( \sqrt{p}+\frac{1}{\sqrt{p}}\right)}^{2}=p+2+\frac{1}{p}=\frac{{p}^{2}+2p+1}{p}=\frac{{\left( 1+p \right)}^{2}}{p}

Therefore

k=2{\left( \sqrt{p}+\frac{1}{\sqrt{p}}\right)}^{2}

Answer: Option 3