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

The Correct Option is C

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, \(\alpha = \frac{4}{3}\).