Question

If one of the zeroes of a quadratic polynomial $(k-1)x^2+kx+1$ is -3, then the value of k is .

• A. $\frac{4}{3}$
• B. $-\frac{4}{3}$
• C. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{4}{3}$

Given $p\left(x\right)=(k-1)x^2+kx+1$ and $p(-3)=0$
$\therefore (k-1)(-3{)}^2+k(-3)+1=0$
$\Rightarrow 9(k-1)-3k+1=0$
$\Rightarrow 9k-9-3k+1=0$
$\Rightarrow 6k-8=0$
$\Rightarrow 6k=8$
$\Rightarrow k=\frac{8}{6}$
$\Rightarrow k=\frac{4}{3}$