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

Step 1: Given the quadratic polynomial:
We are given the quadratic polynomial \( (\alpha - 1)x^2 + \alpha x + 1 \), and it is stated that one of the zeroes of this polynomial is \( -3 \). This means that \( x = -3 \) is a root of the equation. Therefore, we can substitute \( x = -3 \) into the polynomial and set it equal to zero to find the value of \( \alpha \).

Step 2: Substitute \( x = -3 \) into the polynomial:
Substituting \( x = -3 \) into the equation \( (\alpha - 1)x^2 + \alpha x + 1 = 0 \), we get:
\[ (\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0 \] Simplifying the terms:
\[ (\alpha - 1)(9) - 3\alpha + 1 = 0 \] \[ 9(\alpha - 1) - 3\alpha + 1 = 0 \] Distribute the 9:
\[ 9\alpha - 9 - 3\alpha + 1 = 0 \] Now combine like terms:
\[ 9\alpha - 3\alpha - 9 + 1 = 0 \] \[ 6\alpha - 8 = 0 \] Add 8 to both sides:
\[ 6\alpha = 8 \] Now, divide both sides by 6:
\[ \alpha = \frac{8}{6} = \frac{4}{3} \]

Step 3: Conclusion:
The value of \( \alpha \) is \( \frac{4}{3} \).