Question

Find the zeroes of the quadratic polynomial \( 3x^2 - x - 4 \) and verify the relationship between the zeroes and the coefficients.

Hint:

Always verify the sum and product of zeroes using the standard relations \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \).

Answer 1

Step 1: Given polynomial.
\[ p(x) = 3x^2 - x - 4 \] Here, \( a = 3, \; b = -1, \; c = -4 \).

Step 2: Use the quadratic formula to find the zeroes.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the values: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)} = \frac{1 \pm \sqrt{1 + 48}}{6} = \frac{1 \pm \sqrt{49}}{6} \] \[ x = \frac{1 \pm 7}{6} \]
Step 3: Calculate the two roots.
\[ x_1 = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3} \] \[ x_2 = \frac{1 - 7}{6} = \frac{-6}{6} = -1 \]
Step 4: Verify the relationships between the zeroes and coefficients.
Sum of zeroes \( = x_1 + x_2 = \frac{4}{3} - 1 = \frac{1}{3} \).
Product of zeroes \( = x_1 \times x_2 = \frac{4}{3} \times (-1) = -\frac{4}{3} \).
Now, \[ -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}, \quad \frac{c}{a} = \frac{-4}{3} \] Hence verified: \[ \text{Sum of zeroes} = -\frac{b}{a}, \quad \text{Product of zeroes} = \frac{c}{a} \] Step 5: Conclusion.
The zeroes are \( \boxed{\frac{4}{3} \text{ and } -1} \), and the relationship between zeroes and coefficients is verified.