Question

Find the zeroes of the polynomial \(2x^2 + 7x + 5\) and verify the relationship between its zeroes and coefficients.

Hint:

To verify relationships, use sum \(= -\dfrac{b}{a},\) product \(= \dfrac{c}{a}\) from standard quadratic form.

Answer 1

Given Polynomial:
\[ 2x^2 + 7x + 5 = 0 \]

Step 1: Find the zeroes using quadratic formula
For quadratic \(ax^2 + bx + c = 0\), zeroes are:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a=2\), \(b=7\), \(c=5\).
Calculate discriminant:
\[ \Delta = b^2 - 4ac = 7^2 - 4 \times 2 \times 5 = 49 - 40 = 9 \] \[ x = \frac{-7 \pm \sqrt{9}}{2 \times 2} = \frac{-7 \pm 3}{4} \]
Zeroes:
\[ x_1 = \frac{-7 + 3}{4} = \frac{-4}{4} = -1 \] \[ x_2 = \frac{-7 - 3}{4} = \frac{-10}{4} = -\frac{5}{2} \]

Step 2: Verify relationship between zeroes and coefficients
Sum of zeroes:
\[ x_1 + x_2 = -1 - \frac{5}{2} = -\frac{7}{2} \] According to formula:
\[ \text{Sum} = -\frac{b}{a} = -\frac{7}{2} \] Product of zeroes:
\[ x_1 \times x_2 = -1 \times \left(-\frac{5}{2}\right) = \frac{5}{2} \] According to formula:
\[ \text{Product} = \frac{c}{a} = \frac{5}{2} \]

Final Answer:
Zeroes are \(-1\) and \(-\frac{5}{2}\).
Sum of zeroes \(= -\frac{7}{2}\) matches \(-\frac{b}{a}\).
Product of zeroes \(= \frac{5}{2}\) matches \(\frac{c}{a}\).