If the harmonic mean of the roots of 2x2-bx+8-25=0 is 4, then the valueof b =

Let α, β are roots of 2x2−bx+(8+25)=0Given H.M of α, β =4⇒ 2αβα+β=4⇒ 2(8−25)/2(b2)=4⇒ 2(8−25)= 4(bc)⇒ b= (8−25)2⇒ 4−5

If the harmonic mean of the roots of 2x2-bx+8-25=0 is 4, then the valueof b =

Let α, β are roots of 2x2−bx+(8+25)=0Given H.M of α, β =4⇒ 2αβα+β=4⇒ 2(8−25)/2(b2)=4⇒ 2(8−25)= 4(bc)⇒ b= (8−25)2⇒ 4−5

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If the harmonic mean of the roots of $\sqrt{2} x^{2} - bx + (8 - 2 \sqrt{5}) = 0$ is 4, then the value
of b =

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$4 - \sqrt{5}$

$4 + \sqrt{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t α, β a r e r o o t s o f \sqrt{2} x^{2} - b x + (8 + 2 \sqrt{5}) = 0 \\ G i v e n H . M o f α, β = 4 \\ \Rightarrow \frac{2 α β}{α + β} = 4 \\ \Rightarrow \frac{2 (8 - 2 \sqrt{5}) / \sqrt{2}}{(\frac{b}{\sqrt{2}})} = 4 \\ \Rightarrow 2 (8 - 2 \sqrt{5}) = 4 (\frac{b}{c}) \\ \Rightarrow b = \frac{(8 - 2 \sqrt{5})}{2} \\ \Rightarrow 4 - \sqrt{5} \end{array}$