Question

Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)

Hint:

Use quadratic formula and transformation of zeroes formula for new polynomials.

Answer 1

Given polynomial:
\[q(x) = 8x^2 - 2x - 3\]

Step 1: Find the roots of \(q(x)\)
Use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \(a = 8\), \(b = -2\), \(c = -3\).
Calculate the discriminant:
\[\Delta = (-2)^2 - 4 \times 8 \times (-3) = 4 + 96 = 100\] \[x = \frac{2 \pm \sqrt{100}}{2 \times 8} = \frac{2 \pm 10}{16}\]
Roots:
\[x_1 = \frac{2 + 10}{16} = \frac{12}{16} = \frac{3}{4}\] \[x_2 = \frac{2 - 10}{16} = \frac{-8}{16} = -\frac{1}{2}\]

Step 2: Find roots that are 2 less than the roots of \(q(x)\)
New roots:
\[x_1' = \frac{3}{4} - 2 = \frac{3}{4} - \frac{8}{4} = -\frac{5}{4}\] \[x_2' = -\frac{1}{2} - 2 = -\frac{1}{2} - \frac{4}{2} = -\frac{5}{2}\]

Step 3: Form the polynomial with roots \(x_1'\) and \(x_2'\)
Sum of the new roots:
\[S = x_1' + x_2' = -\frac{5}{4} - \frac{5}{2} = -\frac{5}{4} - \frac{10}{4} = -\frac{15}{4}\] Product of the new roots:
\[P = x_1' \times x_2' = \left(-\frac{5}{4}\right) \times \left(-\frac{5}{2}\right) = \frac{25}{8}\]

The polynomial with these roots is:
\[x^2 - Sx + P = 0 \implies x^2 + \frac{15}{4}x + \frac{25}{8} = 0\]
Multiply throughout by 8 to eliminate fractions:
\[8x^2 + 30x + 25 = 0\]

Final Answer:
- Roots of \(q(x)\) are \(\frac{3}{4}\) and \(-\frac{1}{2}\).
- Polynomial with roots 2 less than the roots of \(q(x)\) is:
\[\boxed{8x^2 + 30x + 25 = 0}\]