Question

If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

Hint:

Always try to simplify the required expression in terms of \((\alpha + \beta)\) and \((\alpha\beta)\) first; it saves you from dealing with square roots or complex numbers!

Answer 1

Given:
Polynomial: \[ p(x) = x^2 - 3x - 1 \] Let its zeroes be \(\alpha\) and \(\beta\).

Step 1: Use standard relations of zeroes
For a quadratic \(x^2 + px + q\): \[ \alpha + \beta = -p,\quad \alpha\beta = q \]
Here the polynomial is:
\[ x^2 - 3x - 1 \] So,
\[ \alpha + \beta = 3,\qquad \alpha\beta = -1 \]

Step 2: Find \[ \frac{1}{\alpha} + \frac{1}{\beta} \] Use identity:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \]
Substitute values:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{-1} \] \[ = -3 \]

Final Answer:
\[ \boxed{-3} \]