Question

Find the zeroes of the polynomial \(r(x) = 4x^2 + 3x - 1\). Hence, write a polynomial whose zeroes are reciprocal of the zeroes of \(r(x)\).

Hint:

To get a polynomial with reciprocal roots, take \(x = \frac{1}{\alpha}, \frac{1}{\beta}\) and multiply \( (x - \frac{1}{\alpha})(x - \frac{1}{\beta}) \) or invert roots and form new factors.

Answer 1

Given polynomial:
\[r(x) = 4x^2 + 3x - 1\]

Step 1: Find the zeroes of \(r(x)\)
The zeroes of \(ax^2 + bx + c = 0\) are given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, \(a = 4\), \(b = 3\), and \(c = -1\).
Calculate the discriminant:
\[\Delta = b^2 - 4ac = 3^2 - 4 \times 4 \times (-1) = 9 + 16 = 25\]
Therefore,
\[x = \frac{-3 \pm \sqrt{25}}{2 \times 4} = \frac{-3 \pm 5}{8}\]
The zeroes are:
\[x_1 = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4}\]
\[x_2 = \frac{-3 - 5}{8} = \frac{-8}{8} = -1\]

Step 2: Find polynomial whose zeroes are reciprocals of zeroes of \(r(x)\)
Reciprocal zeroes are:
\[\frac{1}{x_1} = 4, \quad \frac{1}{x_2} = -1\]
Sum of reciprocal zeroes:
\[4 + (-1) = 3\]
Product of reciprocal zeroes:
\[4 \times (-1) = -4\]

The required polynomial is \(p(x) = x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})\)
\[p(x) = x^2 - 3x - 4\]

Final Answer:
Zeroes of \(r(x)\) are \(\frac{1}{4}\) and \(-1\)
Polynomial with zeroes reciprocal to those of \(r(x)\) is:
\[x^2 - 3x - 4 = 0\]