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If the harmonic mean between the roots of $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$ is 4 , then the value of $b$ is
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Verified Answer
The correct answer is:
$4+\sqrt{5}$
Given equation is
$$
(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0
$$
Let $\alpha$ and $\beta$ be the roots of this equation.
$$
\therefore
$$
$$
\begin{aligned}
\alpha+\beta & =\frac{b}{5+\sqrt{2}} \\
\alpha \beta & =\frac{8+2 \sqrt{3}}{5+\sqrt{2}}
\end{aligned}
$$
Given that harmonic mean between the roots of the given equation is 4 .
$$
\begin{array}{rr}
\therefore & \frac{2 \alpha \beta}{\alpha+\beta}=4 \\
\Rightarrow & \frac{8+2 \sqrt{5}}{5+\sqrt{2}} \times \frac{5+\sqrt{2}}{b}=2 \\
\therefore & b=\frac{8+2 \sqrt{5}}{2} \\
& =4+\sqrt{5}
\end{array}
$$
$$
(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0
$$
Let $\alpha$ and $\beta$ be the roots of this equation.
$$
\therefore
$$
$$
\begin{aligned}
\alpha+\beta & =\frac{b}{5+\sqrt{2}} \\
\alpha \beta & =\frac{8+2 \sqrt{3}}{5+\sqrt{2}}
\end{aligned}
$$
Given that harmonic mean between the roots of the given equation is 4 .
$$
\begin{array}{rr}
\therefore & \frac{2 \alpha \beta}{\alpha+\beta}=4 \\
\Rightarrow & \frac{8+2 \sqrt{5}}{5+\sqrt{2}} \times \frac{5+\sqrt{2}}{b}=2 \\
\therefore & b=\frac{8+2 \sqrt{5}}{2} \\
& =4+\sqrt{5}
\end{array}
$$
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