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:

Never try to solve the quadratic equation to find decimal values for \(\alpha\) and \(\beta\).
Expressing symmetric relationships of roots using the sum (\(\alpha+\beta\)) and product (\(\alpha\beta\)) is always faster and prevents working with complicated radical terms!

Answer 1

Step 1: Find the sum and product of the zeroes.
For $p(x) = 3x^2 - 6x - 5$: $\alpha + \beta = -\frac{-6}{3} = 2$ and $\alpha\beta = \frac{-5}{3}$.
Step 2: Rewrite the required expression using the reciprocal sum.
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{\alpha\beta} \]
Step 3: Compute each piece.
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{2}{-5/3} = -\frac{6}{5} \]
So $\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^2 = \frac{36}{25}$, and $\frac{2}{\alpha\beta} = \frac{2}{-5/3} = -\frac{6}{5}$.
Step 4: Combine the pieces.
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{36}{25} - \left(-\frac{6}{5}\right) = \frac{36}{25} + \frac{30}{25} = \frac{66}{25} \]
\[ \boxed{\frac{66}{25}} \]