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

Step 1: Understanding the Concept:
For a quadratic polynomial \(ax^2 + bx + c\), the zeroes \(\alpha\) and \(\beta\) are related to the coefficients. We can find the value of the given expression using these relationships without finding the individual zeroes.
Step 2: Key Formula or Approach:
1. Sum of zeroes: \(\alpha + \beta = -\frac{b}{a}\)
2. Product of zeroes: \(\alpha\beta = \frac{c}{a}\)
3. Identity: \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)
Step 3: Detailed Explanation:
1. For the polynomial \(x^2 - 3x - 1\), we have \(a = 1, b = -3, c = -1\). 2. Calculate sum of zeroes: \[ \alpha + \beta = -\frac{(-3)}{1} = 3 \] 3. Calculate product of zeroes: \[ \alpha\beta = \frac{-1}{1} = -1 \] 4. Substitute these into the expression: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{-1} = -3 \]
Step 4: Final Answer:
The value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is -3.