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

For the quadratic equation $ax^2 + bx + c = 0$,if $\alpha$ and $\beta$ are the roots,then $\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \dots$

If one root of the equation $ax^2 + bx + c = 0$ is the square of the other,then $a(c - b)^3 = cX$,where $X$ is

If the difference of the roots of $x^2 - px + 8 = 0$ is $2$,then the value of $p$ is

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

The roots of the equation $x^2 + bx - c = 0$ where $b, c > 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module