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: Understanding the Question:
The topic of this question is Polynomials.
For a quadratic polynomial \(ax^2 + bx + c\), its zeroes \(\alpha\) and \(\beta\) are related directly to its coefficients.
We are asked to find the value of the symmetric rational expression \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\).
Instead of finding the individual values of \(\alpha\) and \(\beta\) using the quadratic formula, we will express the given fraction in terms of the sum and product of the zeroes.

Step 2: Key Formula or Approach:
- Relate zeroes to coefficients:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha\beta = \frac{c}{a} \] - Simplify the required expression:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \] - Use the algebraic identity for the sum of squares:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] Combine these to get:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} \]

Step 3: Detailed Explanation:

• Identify the coefficients from the polynomial \(3x^2 - 6x - 5\):
Here, \(a = 3\), \(b = -6\), and \(c = -5\).

• Calculate the sum of the zeroes (\(\alpha + \beta\)):
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-6}{3} = 2 \]

• Calculate the product of the zeroes (\(\alpha\beta\)):
\[ \alpha\beta = \frac{c}{a} = \frac{-5}{3} \]

• Write down the expression for the sum of squares (\(\alpha^2 + \beta^2\)):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] \[ \alpha^2 + \beta^2 = (2)^2 - 2\left(-\frac{5}{3}\right) \] \[ \alpha^2 + \beta^2 = 4 + \frac{10}{3} \] \[ \alpha^2 + \beta^2 = \frac{12 + 10}{3} = \frac{22}{3} \]

• Calculate the denominator \((\alpha\beta)^2\):
\[ (\alpha\beta)^2 = \left(-\frac{5}{3}\right)^2 = \frac{25}{9} \]

• Substitute these values into the required expression:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} \] \[ = \frac{\frac{22}{3}}{\frac{25}{9}} \] \[ = \frac{22}{3} \times \frac{9}{25} = \frac{22 \times 3}{25} = \frac{66}{25} \]

Step 4: Final Answer:
The value of \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\) is \(\frac{66}{25}\).