Question

If the equation $\left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0$ has equal roots then
[a] \[{{a}^{2}}\left( 1+{{m}^{2}} \right)\]
[b] \[a\left( 1+{{m}^{2}} \right)\]
[c] \[{{a}^{4}}\left( 1-{{m}^{2}} \right)\]
[d] \[{{a}^{2}}\left( 1-{{m}^{2}} \right)\]

Answer 1

Hint: Assume ‘p’ to be the root of the quadratic expression. Use the formula for the sum of roots and product of roots of the quadratic to get two equations in p. Eliminate p between the equations to get the desired relation after simplification. Alternatively, you can use the property that if a quadratic expression has equal roots, then the discriminant D of the quadratic is 0.

Complete step-by-step answer:

Let p be the root of the quadratic equation $\left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0$.
We know that if m and n are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$, then
$m+n=\dfrac{-b}{a}$ and $mn=\dfrac{c}{a}$
Here we have $a=\left( 1+{{m}^{2}} \right),b=2mc,c={{c}^{2}}-{{a}^{2}},m=p$ and n = p.
Hence we have
$\begin{align}
  & p+p=\dfrac{-2mc}{1+{{m}^{2}}} \\
 & \Rightarrow 2p=\dfrac{-2mc}{1+{{m}^{2}}} \\
 & \Rightarrow p=\dfrac{-mc}{1+{{m}^{2}}}\text{ (i)} \\
\end{align}$
and
$\begin{align}
  & p\times p=\dfrac{{{c}^{2}}-{{a}^{2}}}{1+{{m}^{2}}} \\
 & \Rightarrow {{p}^{2}}=\dfrac{{{c}^{2}}-{{a}^{2}}}{1+{{m}^{2}}} \\
\end{align}$
Substituting the value of p from equation (i), we get
$\begin{align}
  & {{\left( \dfrac{-mc}{1+{{m}^{2}}} \right)}^{2}}=\dfrac{{{c}^{2}}-{{a}^{2}}}{1+{{m}^{2}}} \\
 & \Rightarrow \dfrac{{{m}^{2}}{{c}^{2}}}{{{\left( 1+{{m}^{2}} \right)}^{2}}}=\dfrac{{{c}^{2}}-{{a}^{2}}}{1+{{m}^{2}}} \\
\end{align}$
Cross multiplying, we get
\[\begin{align}
  & {{m}^{2}}{{c}^{2}}\left( 1+{{m}^{2}} \right)=\left( {{c}^{2}}-{{a}^{2}} \right){{\left( 1+{{m}^{2}} \right)}^{2}} \\
 & \Rightarrow {{m}^{2}}{{c}^{2}}=\left( {{c}^{2}}-{{a}^{2}} \right)\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{m}^{2}}{{c}^{2}}={{c}^{2}}+{{m}^{2}}{{c}^{2}}-{{a}^{2}}\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{c}^{2}}-{{a}^{2}}\left( 1+{{m}^{2}} \right)=0 \\
 & \Rightarrow {{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right) \\
\end{align}\]
Hence option [a] is correct.
Note: Alternatively, we know that the roots of the quadratic expression
$a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So if roots are equal, we have
$\begin{align}
  & \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow \sqrt{{{b}^{2}}-4ac}=0 \\
 & \Rightarrow {{b}^{2}}-4ac=0 \\
\end{align}$
Hence a quadratic expression has equal roots if and only if its discriminant = 0.
Hence we have $\left( 1+{{m}^{2}} \right){{x}^{2}}+2mcx+\left( {{c}^{2}}-{{a}^{2}} \right)=0$ if and only if
$\begin{align}
  & {{\left( 2mc \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\
 & \Rightarrow 4{{m}^{2}}{{c}^{2}}-4\left( 1+{{m}^{2}} \right)\left( {{c}^{2}}-{{a}^{2}} \right)=0 \\
 & \Rightarrow {{m}^{2}}{{c}^{2}}=\left( {{c}^{2}}-{{a}^{2}} \right)\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{m}^{2}}{{c}^{2}}={{c}^{2}}+{{m}^{2}}{{c}^{2}}-{{a}^{2}}\left( 1+{{m}^{2}} \right) \\
 & \Rightarrow {{c}^{2}}-{{a}^{2}}\left( 1+{{m}^{2}} \right)=0 \\
 & \Rightarrow {{c}^{2}}={{a}^{2}}\left( 1+{{m}^{2}} \right) \\
\end{align}$
Which is the same as obtained above.
Hence option [a] is correct.