Question

If m, n, k are rational and $m=k+\dfrac{n}{k}$, then the roots of ${{x}^{2}}+mx+n=0$ are:
A. $k,\dfrac{n}{k}$
B. $k,-\dfrac{n}{k}$
C. $-k,-\dfrac{n}{k}$

Answer 1

Hint: We here have been given the equation ${{x}^{2}}+mx+n=0$ and the value of m is being given to us as $m=k+\dfrac{n}{k}$. To find the roots of the given equation, we will first put the value of m given to us in the question into this equation. Then we will use the quadratic formula to find out the roots of this equation which is given for any quadratic equation $a{{x}^{2}}+bx+c=0$ given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Thus, we will put the values of a, b, and c from the equation in this formula, and hence we will obtain the roots of the given equation.

Complete step by step solution:
We here have been given the equation ${{x}^{2}}+mx+n=0$ and we have also been given the value of m in terms of k and n as $m=k+\dfrac{n}{k}$.
To solve this given equation, we will first put the value of m given to us in this equation.
Hence, we will get:
$\begin{align}
  & {{x}^{2}}+mx+n=0 \\
 & \Rightarrow {{x}^{2}}+\left( k+\dfrac{n}{k} \right)x+n=0 \\
\end{align}$
Now, we know that the solutions of any quadratic equation given as $a{{x}^{2}}+bx+c=0$ are given by the quadratic formula given as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here, we can see that:
$\begin{align}
  & a=1 \\
 & b=k+\dfrac{n}{k} \\
 & c=n \\
\end{align}$
Hence, putting the values of a, b and c in the quadratic formula we get:
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \sqrt{{{\left( k+\dfrac{n}{k} \right)}^{2}}-4\left( 1 \right)\left( n \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \sqrt{{{k}^{2}}+\dfrac{{{n}^{2}}}{{{k}^{2}}}+2n-4n}}{2} \\
 & \Rightarrow x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \sqrt{{{k}^{2}}+\dfrac{{{n}^{2}}}{{{k}^{2}}}-2n}}{2} \\
\end{align}$
Now, if we see the terms in the under root sign we will be able to write them as:
$\begin{align}
  & {{k}^{2}}+\dfrac{{{n}^{2}}}{{{k}^{2}}}-2n \\
 & \Rightarrow {{\left( k \right)}^{2}}+{{\left( \dfrac{n}{k} \right)}^{2}}-2\left( k \right)\left( \dfrac{n}{k} \right) \\
\end{align}$
Now, we can see that this is in the form of ${{a}^{2}}+{{b}^{2}}-2ab$ which h is equal to ${{\left( a-b \right)}^{2}}$. Thus, we can write the above mentioned terms as:
$\begin{align}
  & {{\left( k \right)}^{2}}+{{\left( \dfrac{n}{k} \right)}^{2}}-2\left( k \right)\left( \dfrac{n}{k} \right) \\
 & \Rightarrow {{\left( k-\dfrac{n}{k} \right)}^{2}} \\
\end{align}$
Now, putting this back in the quadratic formula we get:
$\begin{align}
  & x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \sqrt{{{k}^{2}}+\dfrac{{{n}^{2}}}{{{k}^{2}}}-2n}}{2} \\
 & x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \sqrt{{{\left( k-\dfrac{n}{k} \right)}^{2}}}}{2} \\
 & \Rightarrow x=\dfrac{-\left( k+\dfrac{n}{k} \right)\pm \left( k-\dfrac{n}{k} \right)}{2} \\
\end{align}$
Thus, we get values of x as:
$\begin{align}
  & x=\dfrac{-\left( k+\dfrac{n}{k} \right)+\left( k-\dfrac{n}{k} \right)}{2},\dfrac{-\left( k+\dfrac{n}{k} \right)-\left( k-\dfrac{n}{k} \right)}{2} \\
&\Rightarrow x=\dfrac{-k-\dfrac{n}{k}+k-\dfrac{n}{k}}{2},\dfrac{-k-\dfrac{n}{k}-k+\dfrac{n}{k}}{2} \\
 & \therefore x=-\dfrac{n}{k},-k \\
\end{align}$
Hence, the roots of the given equation are $-k$ and $-\dfrac{n}{k}$.
Thus, option (C) is the correct option.

Note: After we put the value of m in the given equation, we could have solved it through the method of middle term split also. It is done as follows:
$\begin{align}
  & {{x}^{2}}+\left( k+\dfrac{n}{k} \right)x+n=0 \\
 & \Rightarrow {{x}^{2}}+kx+\dfrac{n}{k}x+n=0 \\
 & \Rightarrow x\left( x+k \right)+\dfrac{n}{k}\left( x+k \right)=0 \\
 & \Rightarrow \left( x+\dfrac{n}{k} \right)\left( x+k \right)=0 \\
\end{align}$
Thus, we get the values of x as:
$\begin{align}
  & x+\dfrac{n}{k}=0 \\
 & \therefore x=-\dfrac{n}{k} \\
 & x+k=0 \\
 & \therefore x=-k \\
\end{align}$