Question

Let ${{x}^{2}}-2ax+{{b}^{2}}=0$ and ${{x}^{2}}-2bx+{{a}^{2}}=0$ be two equations. Then the AM of the roots of the first equation is
(a) AM of the roots of the second
(b) GM of the roots of the second
(c) square root of the GM of the roots of the second
(d) None of these

Answer 1

Hint: We have to firstly, consider equation ${{x}^{2}}-2ax+{{b}^{2}}=0$ and apply the properties of the roots of quadratic equation, that is, the sum of the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to $-\dfrac{b}{a}$ . Using this property, we will get the AM of the first equation. Now, we have to consider the second equation. We have to apply the property of the product of the roots of the quadratic equation. From this property, we will get the required solution.

Complete step-by-step solution:
We are given with ${{x}^{2}}-2ax+{{b}^{2}}=0$ . Let us assume the roots of this equation as $\alpha $ and $\beta $ . We know that for a quadratic equation, $a{{x}^{2}}+bx+c=0$ with roots $p$ and $q$ , the sum of the roots is equal to $-\dfrac{b}{a}$ .
$\Rightarrow p+q=\dfrac{-b}{a}$
Here, for the first equation, we can see that $a=1,b=-2a\text{ and }c={{b}^{2}}$ .
$\begin{align}
  & \Rightarrow \alpha +\beta =-\dfrac{\left( -2a \right)}{1} \\
 & \Rightarrow \alpha +\beta =2a \\
\end{align}$
Let us take 2 to the LHS.
$\Rightarrow a=\dfrac{\alpha +\beta }{2}...\left( i \right)$
We know that arithmetic mean (AM) of n values, say, ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is the sum of these values divided by the total number of values (that is, n). Therefore, from equation (i), we can say that the arithmetic mean of the roots of the first equation is equal to a.
We know that for a quadratic equation, $a{{x}^{2}}+bx+c=0$ , the product of the roots p and q is equal to $\dfrac{c}{a}$ .
$\Rightarrow pq=\dfrac{c}{a}$
For the first equation, we can write
$\begin{align}
  & \Rightarrow \alpha \beta =\dfrac{{{b}^{2}}}{1} \\
 & \Rightarrow \alpha \beta ={{b}^{2}} \\
\end{align}$
Now, let us consider the second equation ${{x}^{2}}-2bx+{{a}^{2}}=0$ . We can assume the roots of this equation as $\alpha '$ and $\beta '$ . When we compare this equation to the standard form $a{{x}^{2}}+bx+c=0$ , we can see that $a=1,b=-2b\text{ and }c={{a}^{2}}$ . Then, according to the properties of the roots of the quadratic polynomial, we can write the sum of the roots as
$\begin{align}
  & \Rightarrow \alpha '+\beta '=-\dfrac{\left( -2b \right)}{1} \\
 & \Rightarrow \alpha '+\beta '=2b...\left( ii \right) \\
\end{align}$
We can also write the product of the roots as
$\begin{align}
  & \Rightarrow \alpha '\beta '=\dfrac{{{a}^{2}}}{1} \\
 & \Rightarrow \alpha '\beta '={{a}^{2}} \\
\end{align}$
Let us take the square root on both the sides in the above equation.
$\Rightarrow a=\sqrt{\alpha '\beta '}...\left( iii \right)$
We know that geometric mean (GM) of n observations, say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is the ${{n}^{th}}$ root of the product of these observations. Therefore, from equation (iii), we can see that the GM of the roots of the second equation is equal to a.
From equations (i) and (iii), we can write
$\Rightarrow a=\dfrac{\alpha +\beta }{2}=\sqrt{\alpha '\beta '}$
Therefore, we can conclude that the AM of the roots of the first equation is the GM of the roots of the second equation.
Hence, the correct option is b.

Note: Students must thoroughly understand the properties of the roots of the polynomials. The main property includes the sum of roots, product of roots and sum of product of the roots taken two at a time. The last property will be present of cubic polynomials and above. Students must know the formula for AM and GM. AM of n observations, say, ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is given by
$\Rightarrow AM=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}$
GM of n observations, say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is given by
$\Rightarrow GM=\sqrt[n]{{{x}_{1}}{{x}_{2}}...{{x}_{n}}}$