Question

The harmonic mean of the roots of the equation $(5+\sqrt{2}){\mathrm{x}}^2-(4+\sqrt{5})\mathrm{x}+8+2\sqrt{5}=0$ is

Answer 1

Given:

$(5+\sqrt{2}){\mathrm{x}}^2-(4+\sqrt{5})\mathrm{x}+\left(8+2\sqrt{5}\right)=0$

It is a quadratic equation of the form $\mathrm{a}{\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0$

$\mathrm{a}=5+\sqrt{2},\mathrm{b}=-\left(4+\sqrt{5}\right)\mathrm{c}=\left(8+2\sqrt{5}\right)$

Let $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of the given quadratic equation.

Then the sum of roots will be $\mathrm{\alpha }+\mathrm{\beta }=$ $\frac{-\mathrm{b}}{\mathrm{a}}$ and the product of roots is $\mathrm{\alpha \beta }=$$\frac{\mathrm{c}}{\mathrm{a}}$

$\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=\frac{4+\sqrt{5}}{5+\sqrt{2}},\mathrm{\alpha \beta }=\frac{8+2\sqrt{5}}{5+\sqrt{2}}$

The harmonic mean of two numbers $\mathrm{\alpha }$ and $\mathrm{\beta }$ will be

$\begin{array}{rcl}\mathrm{HM}& =& \frac{2\mathrm{\alpha \beta }}{\mathrm{\alpha }+\mathrm{\beta }}\\ & =& \frac{2{\displaystyle \left(\frac{{\displaystyle 8+2\sqrt{5}}}{{\displaystyle 5+\sqrt{2}}}\right)}}{{\displaystyle \left(\frac{{\displaystyle 4+\sqrt{5}}}{{\displaystyle 5+\sqrt{2}}}\right)}}\\ & =& \frac{2\left(8+2\sqrt{5}\right)}{4+\sqrt{5}}\\ & =& \frac{4\left(4+\sqrt{5}\right)}{4+\sqrt{5}}\\ & =& 4\end{array}$

Hence, the harmonic mean of the roots of the equation $(5+\sqrt{2}){\mathrm{x}}^2-(4+\sqrt{5})\mathrm{x}+8+2\sqrt{5}=0$ is $4$.