Question

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

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

Answer 1

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

According to the question,
-3 is one of the zeros of quadratic polynomial $(k–1)x^2+kx+1.$
Substituting -3 in the given polynomial,
$(k–1)(-3{)}^2+k(-3)+1=0$
$(k–1)9-3k+1=0$
$9k-9-3k+1=0$
$6k-8=0$
$6k=8$
$k=\frac{8}{6}$
$k=\frac{4}{3}$