Question

Find the H.M. of two roots in the equation ${{x}^{2}}-10x-11=0$\[\]
A.$\dfrac{1}{5}$\[\]
B. $\dfrac{5}{21}$\[\]
C. $\dfrac{21}{20}$\[\]
D. $\dfrac{11}{5}$\[\]

Answer 1

Hint: We compare the given quadratic equation with general quadratic equation $a{{x}^{2}}+bx+c=0,a\ne 0$ and find the sum of the two roots (say $\alpha ,\beta $ ) of with the formula $\alpha +\beta =\dfrac{-b}{a}$ and then product of the roots $\alpha \beta =\dfrac{c}{a}$. We put these values in the formula of harmonic mean for two values HM$=\dfrac{2\alpha \beta }{\alpha +\beta }$\[\]

Complete step by step answer:
We know that harmonic progression is a sequence of reciprocals of the terms occurring in any arithmetic sequence. We also know that harmonic means abbreviated as H.M. is the reciprocal of arithmetic mean of reciprocals of the observation. If there are $n$ number of observation say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ where none of the observation is zero then their reciprocals are $\dfrac{1}{{{x}_{1}}},\dfrac{1}{{{x}_{2}}},...,\dfrac{1}{{{x}_{n}}}$. The arithmetic mean $m$ of reciprocals is
\[m=\dfrac{\dfrac{1}{{{x}_{1}}}+\dfrac{1}{{{x}_{2}}}+...+\dfrac{1}{{{x}_{n}}}}{n}=\dfrac{\sum\limits_{i=1}^{n}{\dfrac{1}{{{x}_{i}}}}}{n}\]
The harmonic mean is the inverse of the arithmetic mean of the reciprocals. So we have
 \[\text{HM}={{\left( \dfrac{\sum\limits_{i=1}^{n}{\dfrac{1}{{{x}_{i}}}}}{n} \right)}^{-1}}=\dfrac{n}{\dfrac{1}{{{x}_{1}}}+\dfrac{1}{{{x}_{2}}}+...+\dfrac{1}{{{x}_{n}}}}\]

We get the harmonic mean for two numbers by putting $n=2$ in the above formula and get as,
\[\text{HM}=\dfrac{2}{\dfrac{1}{{{x}_{1}}}+\dfrac{1}{{{x}_{2}}}}=\dfrac{2{{x}_{1}}{{x}_{2}}}{{{x}_{1}}+{{x}_{2}}}\]
The given equation is
${{x}^{2}}-10x+11=0$
We see that the given equation is a quadratic equation of the type $a{{x}^{2}}+bx+c=0$ which has roots say ${{x}_{1}},{{x}_{2}}$ then we know that the sum of the roots is given by ${{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}$ and the product of the roots is give by ${{x}_{1}}{{x}_{2}}=\dfrac{c}{a}$. \[\]
We compare the given equation with the general quadratic equation $a{{x}^{2}}+bx+c=0$ and find $a=1$,$b=-10$ and $c=11$. So the sum of the roots is ${{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}=\dfrac{-\left( -10 \right)}{1}=10$ and the product of the roots is ${{x}_{1}}{{x}_{2}}=\dfrac{c}{a}=\dfrac{11}{1}=11$. We put these values in the formula for harmonic mean and get

\[\text{HM}=\dfrac{2{{x}_{1}}{{x}_{2}}}{{{x}_{1}}+{{x}_{2}}}=\dfrac{2\left( 10 \right)}{11}=\dfrac{5}{6}\]
So the correct option is D. \[\]

Note:
The harmonic mean of any three numbers is given by $\dfrac{3{{x}_{1}}{{x}_{2}}{{x}_{3}}}{{{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{1}}+{{x}_{1}}{{x}_{3}}}$. If they are roots of sum cubic equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$ then we have product of ${{x}_{1}}{{x}_{2}}{{x}_{3}}=\dfrac{-d}{a}$ and sum of products of two roots ${{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{1}}+{{x}_{1}}{{x}_{3}}=\dfrac{c}{a}$