The harmonic mean of the roots of equation $(4+\sqrt{5})x^2 - (2+\sqrt{3})x + 6+3\sqrt{3} = 0$ will be -

Explanation

Let α, β be the root of the equation ∴ Sum of roots = $\alpha + \beta = -\frac{b}{a}$ $$ = \frac{2+\sqrt{3}}{4+\sqrt{5}} $$ Product of roots = $\alpha.\beta = \frac{c}{a}$ $$ = \frac{6+3\sqrt{3}}{4+\sqrt{5}} = \frac{3(2+\sqrt{3})}{4+\sqrt{5}} $$ ∴ The harmonic mean of roots is = $\frac{2\alpha\beta}{\alpha+\beta}$ $$ = \frac{\frac{2 \times 3(2)+\sqrt{3}}{4+\sqrt{5}}}{\left(\frac{2+\sqrt{3}}{4+\sqrt{5}}\right)} = 6 $$ ## Page 29