Q52 / 160 · AP EAMCET 2015 · Mathematics

If the harmonic mean of the roots $\sqrt{2} x^2-b x+(8-2 \sqrt{d})=0$ is 4 , then the value of $b$ is

  • A
    $2$
  • B
    $3$
  • Correct answer
    C
    $4-\sqrt{5}$
  • D
    $4+\sqrt{5}$

Solution

Given equation $\sqrt{2} x^2-b x+(8-2 \sqrt{5})=0$ Let the roots of the given equation are $\alpha$ and $\beta$. Sum of roots, $\alpha+\beta=\frac{-b}{a}=-\left(\frac{-b}{\sqrt{2}}\right)=\frac{b}{\sqrt{2}}$ and product of roots, $\alpha \beta=\frac{c}{a}=\frac{8-2 \sqrt{5}}{\sqrt{2}}$ Now, as given in the question $\begin{aligned} & \frac{2 \alpha \beta}{\alpha+\beta}=4 \Rightarrow \frac{2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)}{\frac{b}{\sqrt{2}}}=4 \\ & \Rightarrow \quad 2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)=b \Rightarrow b=4-\sqrt{5} \end{aligned}$

Hence, the correct answer is option: 'C'.

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