Question

If one of the zeros of the quadratic polynomial $\left(k-1\right)x^2+kx+1$ is $-3$, then find the value of $k$.

Answer 1

Find the value of $k$.

Since, the roots of a quadratic polynomial satisfy the polynomial.

$$\begin{gathered} \left(k-1\right){\left(-3\right)}^2+k\left(-3\right)+1=0 \\ \Rightarrow \left(k-1\right)9-3k+1=0 \\ \Rightarrow 9k-9-3k+1=0 \\ \Rightarrow 6k=8 \\ \Rightarrow k=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$

Hence, the value of $k$ is $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$.