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:
We are given a quadratic polynomial:
\[ (\alpha - 1)x^2 + \alpha x + 1 \]
It is also given that one of the zeroes of this polynomial is \(x = -3\). We are to find the value of \(\alpha\).

Step 1: Substitute the known root into the polynomial
Since \(-3\) is a zero of the polynomial, it must satisfy the equation:
\[ (\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0 \]
Now simplify step-by-step:
\[ (\alpha - 1)(9) - 3\alpha + 1 = 0 \]
\[ 9\alpha - 9 - 3\alpha + 1 = 0 \]
\[ (9\alpha - 3\alpha) + (-9 + 1) = 0 \Rightarrow 6\alpha - 8 = 0 \]

Step 2: Solve for \(\alpha\)
\[ 6\alpha = 8 \Rightarrow \alpha = \frac{8}{6} = \frac{4}{3} \]

Final Answer:
The value of \(\alpha\) is \(\frac{4}{3}\).