If the harmonic mean of the roots of the equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the roots of the quadratic equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equation

Vedclass · Accepted Answer

$4 - \sqrt{5}$

Vedclass · Answer

(C) Let the roots of the quadratic equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ be $\alpha$ and $\beta$.From the properties of quadratic equations,the sum of roots $\alpha + \beta = \frac{b}{\sqrt{2}}$ and the product of roots $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}} = 4\sqrt{2} - \sqrt{10}$.The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.Given $HM = 4$,we have $4 = \frac{2(4\sqrt{2} - \sqrt{10})}{\frac{b}{\sqrt{2}}}$.$4 = \frac{2(4\sqrt{2} - \sqrt{10}) \cdot \sqrt{2}}{b}$.$4 = \frac{2(8 - \sqrt{20})}{b} = \frac{2(8 - 2\sqrt{5})}{b} = \frac{16 - 4\sqrt{5}}{b}$.$4b = 16 - 4\sqrt{5}$.Dividing by $4$,we get $b = 4 - \sqrt{5}$.

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

Similar Questions

Find $\alpha^4+\beta^4$ if $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.

If the roots of $x^2 - 7x + 6 = 0$ are $\alpha$ and $\beta$,then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-3x^2+3x+1=0$,then $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=$

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c$,where $c \neq 0$,then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

The $G.M.$ of the roots of the equation $x^2 - 18x + 9 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module