If the harmonic mean between the roots of (5+ | vedclass

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

Question

(D) 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.From the relation between roots a

Vedclass · Accepted Answer

$4+\sqrt{5}$

Vedclass · Answer

(D) 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.From the relation between roots and coefficients:$\alpha+\beta = \frac{b}{5+\sqrt{2}}$$\alpha \beta = \frac{8+2 \sqrt{5}}{5+\sqrt{2}}$The harmonic mean $(HM)$ between the roots is given by $HM = \frac{2 \alpha \beta}{\alpha+\beta}$.Given $HM = 4$,we have:$\frac{2 \alpha \beta}{\alpha+\beta} = 4$Substituting the values of $\alpha+\beta$ and $\alpha \beta$:$\frac{2 	imes \frac{8+2 \sqrt{5}}{5+\sqrt{2}}}{\frac{b}{5+\sqrt{2}}} = 4$$\frac{2(8+2 \sqrt{5})}{b} = 4$$\frac{8+2 \sqrt{5}}{b} = 2$$b = \frac{8+2 \sqrt{5}}{2} = 4+\sqrt{5}$.

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

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3 - 9x^2 + 5x - 7 = 0$,then what is the value of $\alpha + \beta + \gamma$?

If the roots of the equation $ax^2 + bx + c = 0$ are real and of the form $\frac{\alpha}{\alpha - 1}$ and $\frac{\alpha + 1}{\alpha}$,then the value of $(a + b + c)^2$ is

If the difference between the roots of the equation $x^2 + px + q = 0$ is $1$,then:

If one root of $x^3-7x^2+36=0$ is twice the other,then the sum of those two roots is

Vedclass Test Series

Exam Paper Generator

Online Exam Module