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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
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Verified Answer
The correct answer is:
$4-\sqrt{5}$
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
$\begin{aligned}
& \frac{2 \alpha \beta}{\alpha+\beta}=4 \Rightarrow \frac{2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)}{\frac{b}{\sqrt{2}}}=4 \\
& \Rightarrow \quad 2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)=b \Rightarrow b=4-\sqrt{5}
\end{aligned}$
$\sqrt{2} x^2-b x+(8-2 \sqrt{5})=0$
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
$\begin{aligned}
& \frac{2 \alpha \beta}{\alpha+\beta}=4 \Rightarrow \frac{2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)}{\frac{b}{\sqrt{2}}}=4 \\
& \Rightarrow \quad 2\left(\frac{8-2 \sqrt{5}}{\sqrt{2}}\right)=b \Rightarrow b=4-\sqrt{5}
\end{aligned}$
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