Question

If one of the zeros of the quadratic polynomial $\left( k-1 \right){{x}^{2}}+kx+1$ is $-3$, then the value of $k$ is
A. $\dfrac{4}{3}$
B. $\dfrac{-4}{3}$
C. $\dfrac{2}{3}$
D. $\dfrac{-2}{3}$

Answer 1

Hint: First we will assume another root of the given equation as $\alpha $. We have the relation between $p,q$ which are the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ and the coefficients of the quadratic equation $a{{x}^{2}}+bx+c=0$ as
$p+q=\dfrac{-b}{a}$
$pq=\dfrac{c}{a}$
From the above relation we will make an equation to find the value of $k$.

Complete step-by-step solution
Given that, the quadratic equation is
$\left( k-1 \right){{x}^{2}}+kx+1=0$
Comparing the above equation with $a{{x}^{2}}+bx+c=0$, we get $a=k-1$, $b=k$, $c=1$
Given that the value of one root is $-3$, then $p=-3$
Let the other root of the equation is $\alpha $, then $q=\alpha $
We know that $p+q=\dfrac{-b}{a}$
Substituting the values of $p=-3$, $q=\alpha $,$b=k$, $a=k-1$ in the above equation, we get
$\begin{align}
  & -3+\alpha =\dfrac{-k}{k-1} \\
 & \Rightarrow \left( -3+\alpha \right)\left( k-1 \right)=-k \\
 & \Rightarrow -3k+3+\alpha k-\alpha +k=0 \\
 & \Rightarrow k\left( \alpha -2 \right)+3-\alpha =0 \\
 & \Rightarrow k\left( \alpha -2 \right)=\alpha -3 \\
 & \Rightarrow k=\dfrac{\alpha -3}{\alpha -2}...\left( \text{i} \right) \\
\end{align}$
We know that $pq=\dfrac{c}{a}$
Substituting the values of $p=-3$,$q=\alpha $,$a=k-1$,$c=1$, we get
$\begin{align}
  & -3\alpha =\dfrac{1}{k-1} \\
 & \Rightarrow \alpha =\dfrac{1}{3\left( 1-k \right)} \\
\end{align}$
Substituting the value of $\alpha $ from the above equation in equation $\left( \text{i} \right)$, we get
$\begin{align}
  & k=\dfrac{\alpha -3}{\alpha -2} \\
 & \Rightarrow k=\dfrac{\dfrac{1}{3\left( 1-k \right)}-3}{\dfrac{1}{3\left( 1-k \right)}-2} \\
 & \Rightarrow k=\dfrac{\dfrac{1-9\left( 1-k \right)}{3\left( 1-k \right)}}{\dfrac{1-6\left( 1-k \right)}{3\left( 1-k \right)}} \\
\end{align}$
Now cancelling the term $3\left( 1-k \right)$ in both numerator and denominator, we get
$k=\dfrac{1-9+9k}{1-6+6k}$
Now multiplying $6k-5$ on both sides, we get
$\begin{align}
  & \Rightarrow k\left( 6k-5 \right)=\dfrac{9k-8}{6k-5}\left( 6k-5 \right) \\
 & \Rightarrow 6{{k}^{2}}-5k=9k-8 \\
\end{align}$
Now adding the term $8-9k$ to both sides of the above equation, we get
$\begin{align}
  & \Rightarrow 6{{k}^{2}}-5k+8-9k=9k-8+8-9k \\
 & \Rightarrow 6{{k}^{2}}-14k+8=0 \\
\end{align}$
Now dividing the above equation with $2$, we get
$\begin{align}
  & \Rightarrow \dfrac{1}{2}\left( 6{{k}^{2}}-14k+8 \right)=0\times \dfrac{1}{2} \\
 & \Rightarrow 3{{k}^{2}}-7k+4=0 \\
\end{align}$
Solving the above equation to get the value of $k$, we get
$\Rightarrow 3{{k}^{2}}-3k-4k+4=0$
Taking $3k$common from $3{{k}^{2}}-3k$ and $-4$ from $-4k+4$ , we get
$\begin{align}
  & \Rightarrow 3{{k}^{2}}-3k-4k+4=0 \\
 & \Rightarrow 3k\left( k-1 \right)-4\left( k-1 \right)=0 \\
\end{align}$
Now taking $\left( k-1 \right)$ common from the above equation, we get
$\Rightarrow \left( k-1 \right)\left( 3k-4 \right)=0$
Now equating first term to zero to get the value of $k$, then we will have
$\begin{align}
  & \Rightarrow k-1=0 \\
 & \therefore k=1 \\
\end{align}$
 Now equating the second term to zero to get another value of $k$, then we will have
 $\begin{align}
  & \Rightarrow 3k-4=0 \\
 & \Rightarrow 3k=4 \\
 & \therefore k=\dfrac{4}{3} \\
\end{align}$
$\therefore $ the values of $k$ are $k=1\text{ or }\dfrac{4}{3}$
Hence the answer is $k=1\text{ or }\dfrac{4}{3}$.

Note: We can also solve the above equation by simple method. i.e. substituting the zero of the equation $-3$ in the equation $\left( k-1 \right){{x}^{2}}+kx+1$, we get
$\begin{align}
  & \left( k-1 \right){{\left( -3 \right)}^{2}}+k\left( -3 \right)+1=0 \\
 & \Rightarrow 9\left( k-1 \right)-3k+1=0 \\
 & \Rightarrow 9k-9-3k+1=0 \\
 & \Rightarrow 6k=8 \\
 & \therefore k=\dfrac{4}{3} \\
\end{align}$
$\therefore $ From both the methods we got the same answer.