Question

\(\alpha, \beta\) are zeroes of the polynomial \(p(x) = 3x^2 - 6x - 5\). Find the value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\).

Hint:

Avoid solving for \(\alpha\) and \(\beta\) individually using the quadratic formula unless the polynomial factors easily. Always aim to manipulate the expression into forms of \((\alpha+\beta)\) and \((\alpha\beta)\).

Answer 1

Step 1: Understanding the Concept:
The relationship between zeroes and coefficients of a quadratic polynomial \(ax^2 + bx + c\) is:
Sum of zeroes (\(\alpha + \beta\)) = \(-\frac{b}{a}\).
Product of zeroes (\(\alpha \beta\)) = \(\frac{c}{a}\).
Step 2: Key Formula or Approach:
We need to evaluate \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2}\).
Use the identity: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\).
Step 3: Detailed Explanation:
For \(p(x) = 3x^2 - 6x - 5\):
\(a = 3, b = -6, c = -5\).
\[ \alpha + \beta = -\frac{-6}{3} = 2 \]
\[ \alpha \beta = \frac{-5}{3} \]
Now, find \(\alpha^2 + \beta^2\):
\[ \alpha^2 + \beta^2 = (2)^2 - 2\left(\frac{-5}{3}\right) = 4 + \frac{10}{3} = \frac{12 + 10}{3} = \frac{22}{3} \]
Now, calculate the required expression:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{\frac{22}{3}}{\left(-\frac{5}{3}\right)^2} \]
\[ = \frac{\frac{22}{3}}{\frac{25}{9}} = \frac{22}{3} \times \frac{9}{25} \]
\[ = \frac{22 \times 3}{25} = \frac{66}{25} \]
Step 4: Final Answer:
The value is \(\frac{66}{25}\).