Question

The zeroes of quadratic polynomial $6x^2 - 7x - 3$ will be:

• A. $-\tfrac{3}{2}, -\tfrac{1}{3}$
• B. $\tfrac{3}{2}, -\tfrac{1}{3}$
• C. $-\tfrac{3}{2}, \tfrac{1}{3}$
• D. $-9, 2$

Hint:

Always apply the quadratic formula carefully. Check signs of $b$ and $c$ to avoid errors.

Answer 1

The Correct Option is B

Step 1: Write quadratic polynomial in standard form
\[ 6x^2 - 7x - 3 = 0 \] Here $a = 6$, $b = -7$, $c = -3$.

Step 2: Apply quadratic formula
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{2(6)} \] \[ x = \frac{7 \pm \sqrt{49 + 72}}{12} \] \[ x = \frac{7 \pm \sqrt{121}}{12} \] \[ x = \frac{7 \pm 11}{12} \]

Step 3: Simplify roots
Case 1: $\dfrac{7 + 11}{12} = \dfrac{18}{12} = \dfrac{3}{2}$
Case 2: $\dfrac{7 - 11}{12} = \dfrac{-4}{12} = -\dfrac{1}{3}$

Step 4: Conclusion
The roots are $\tfrac{3}{2}$ and $-\tfrac{1}{3}$.
The correct answer is option (B).