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

Problem:
Given the quadratic polynomial \((\alpha - 1)x^2 + \alpha x + 1\), and knowing that \(x = -3\) is one of its zeroes, determine the value of \(\alpha\).

Solution:
Since \(x = -3\) is a zero, it must satisfy the polynomial equation:
\[ (\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0 \] Simplify the equation:
\[ (\alpha - 1)(9) - 3\alpha + 1 = 0 \] \[ 9\alpha - 9 - 3\alpha + 1 = 0 \] Combine like terms:
\[ (9\alpha - 3\alpha) + (-9 + 1) = 0 \] \[ 6\alpha - 8 = 0 \] Solve for \(\alpha\):
\[ 6\alpha = 8 \] \[ \alpha = \frac{8}{6} \] \[ \alpha = \frac{4}{3} \]
Final Answer:
The value of \(\alpha\) is \(\frac{4}{3}\).