Question

If one of the zeroes of the quadratic polynomial $(\alpha - 1)x^2 + \alpha x + 1$ is $-3$, then the value of $\alpha$ is:

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

Answer 1

Answer: (C)

$\frac{4}{3}$

Answer 2

Let the quadratic polynomial be $f(x) = (\alpha - 1)x^2 + \alpha x + 1$.

Given that one of the zeroes is $-3$, we can substitute $x = -3$ into the equation:

$f(-3) = (\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0$

Simplifying:

$(\alpha - 1)(9) - 3\alpha + 1 = 0$

$9\alpha - 9 - 3\alpha + 1 = 0$

$6\alpha - 8 = 0$

Solve for $\alpha$:

$6\alpha = 8 \implies \alpha = \frac{8}{6} = \frac{4}{3}$

Thus, $\alpha = \frac{4}{3}$.