Question

If the harmonic mean of the roots of $\sqrt{2}{{x}^{2}}-bx+(8-2\sqrt{5})=0$ is 4, then the value of b is
(a) 2
(b) 3
(c) $4-\sqrt{5}$
(d) $4+\sqrt{5}$

Answer 1

Hint: As this is a quadratic equation, it will have two roots and we can use their sum and multiplication results and the formula for the harmonic mean to obtain the required answer.
Complete step-by-step answer:
Let $a{{x}^{2}}+bx+c=0$ be a quadratic equation. Then, the sum and product of the roots (say $\alpha$ and $\beta$) are given by
$\alpha +\beta =-\dfrac{coefficient\text{ }of\text{ }x}{coefficient\text{ }of\text{ }{{x}^{2}}}$ and $\alpha \beta =\dfrac{\text{constant term}}{\text{coefficient of }{{x}^{2}}}$ respectively i.e
$\alpha +\beta =-\dfrac{b}{a}$ and $\alpha \beta =\dfrac{c}{a}.........(1.1)$
It is given that$\sqrt{2}{{x}^{2}}-bx+(8-2\sqrt{5})=0$.
Let the roots of this Equation be $\alpha$ and $\beta$. Then by equation (1.1), their sum and product are given by
$\alpha +\beta =\dfrac{-(-b)}{\sqrt{2}}=\dfrac{b}{2}$ and $\alpha \beta =\dfrac{8-2\sqrt{5}}{\sqrt{2}}....(1.2)$.
The harmonic mean of n numbers (k1, k2 … kn) is given by
Harmonic Mean of k1, k2 … kn =$\dfrac{n}{\dfrac{1}{{{k}_{1}}}+\dfrac{1}{{{k}_{2}}}+...+\dfrac{1}{{{k}_{n}}}}\ldots \ldots \ldots (1.3)$
Now, it is given that the Harmonic Mean of the roots is 4. Thus, by taking ${{k}_{1}}=\alpha \text{ and }{{k}_{2}}=\beta$ and n=2 in equation (1.2), we obtain
$\dfrac{2}{\dfrac{1}{\alpha }+\dfrac{1}{\beta }}=4$
$\Rightarrow \dfrac{2}{\dfrac{\alpha +\beta }{\alpha \beta }}=4$
$\dfrac{2\alpha \beta }{\alpha +\beta }=4$…………………$(1.4)$
Putting the obtained values of the sum and product of the roots (from equation (1.2)) in equation (1.3), we get,
$\dfrac{2\left( \dfrac{8-2\sqrt{5}}{\sqrt{2}} \right)}{\dfrac{b}{\sqrt{2}}}=\dfrac{2(8-2\sqrt{5})}{b}=4$
$\Rightarrow b=\dfrac{8-2\sqrt{5}}{2}$
$\Rightarrow b=4-\sqrt{5}$
We get the value of b to be $4-\sqrt{5}$. Thus, the correct answer of the question is option (c) $4-\sqrt{5}$.

Note: One should be careful while taking the harmonic mean as it is given by the reciprocals of the numbers in the denominator and not just the inverse of the normal mean $\left( \dfrac{{{k}_{1}}+{{k}_{2}}+...+{{k}_{n}}}{n} \right)$ of the numbers.