IPMAT Indore 2023 · Quantitative Ability · Question 39

Step 1 — write the harmonic mean of two numbers. For roots \alpha and \beta:

HM=\frac{2}{\frac{1}{\alpha}+\frac{1}{\beta}}=\frac{2\alpha \beta}{\alpha +\beta}

Step 2 — bring in Vieta’s formulas. For A{x}^{2}+Bx+C=0:

\alpha +\beta =-\frac{B}{A}=\frac{b}{5+\sqrt{2}}, \alpha \beta =\frac{C}{A}=\frac{8+2\sqrt{5}}{5+\sqrt{2}}

Step 3 — substitute. The common denominator \left( 5+\sqrt{2}\right) cancels completely:

HM=\frac{2\cdot \frac{8+2\sqrt{5}}{5+\sqrt{2}}}{\frac{b}{5+\sqrt{2}}}=\frac{2\left( 8+2\sqrt{5}\right)}{b}

Step 4 — set it to 4 and solve.

\frac{2\left( 8+2\sqrt{5}\right)}{b}=4 \Rightarrow b=\frac{16+4\sqrt{5}}{4}=4+\sqrt{5}

Answer: Option 2 — 4+\sqrt{5}

Shortcut. The leading coefficient always cancels — so ignore it. For A{x}^{2}+Bx+C=0,

HM=\frac{2\alpha \beta}{\alpha +\beta}=\frac{2C/A}{-B/A}=\frac{2C}{-B}

so the harmonic mean depends only on B and C. Here -B=b and C=8+2\sqrt{5}:

4=\frac{2\left( 8+2\sqrt{5}\right)}{b}\Rightarrow b=\frac{8+2\sqrt{5}}{2}=4+\sqrt{5}

One division, and the \left( 5+\sqrt{2}\right) never enters — it is there purely to look forbidding and to tempt you into rationalising a denominator you never needed.

Worth memorising alongside it: AM=-B/2A and GM=\sqrt{C/A}. Each picks out a different pair of coefficients.