Question

If \(\alpha, \beta\) are zeroes of the polynomial \(8x^2 - 5x - 1\), then form a quadratic polynomial in x whose zeroes are \(\frac{2}{\alpha}\) and \(\frac{2}{\beta}\).

Answer 1

Transforming a Polynomial Using Modified Zeroes

Given polynomial:

\[ P(x) = 8x^2 - 5x - 1 \]

Step 1: Identify Original Zeroes

If \( \alpha \) and \( \beta \) are the zeroes of the polynomial, then:

• Sum: \[ \alpha + \beta = \frac{-(-5)}{8} = \frac{5}{8} \]
• Product: \[ \alpha \beta = \frac{-1}{8} \]

Step 2: Define New Zeroes

Let the new zeroes be:

\[ \alpha' = \frac{2}{\alpha}, \quad \beta' = \frac{2}{\beta} \]

Step 3: Calculate Sum and Product of New Zeroes

Sum:

\[ \alpha' + \beta' = \frac{2}{\alpha} + \frac{2}{\beta} = 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = 2\left(\frac{\alpha + \beta}{\alpha \beta}\right) = 2 \left( \frac{5/8}{-1/8} \right) = 2 \times (-5) = -10 \]

Product:

\[ \alpha' \cdot \beta' = \frac{2}{\alpha} \cdot \frac{2}{\beta} = \frac{4}{\alpha \beta} = \frac{4}{-1/8} = -32 \]

Step 4: Form the New Quadratic Polynomial

A quadratic polynomial with zeroes \( \alpha' \) and \( \beta' \) is: \[ x^2 - (\alpha' + \beta')x + \alpha'\beta' \] Substituting: \[ x^2 - (-10)x + (-32) = x^2 + 10x - 32 \]

✅ Final Answer:

\[ \boxed{x^2 + 10x - 32} \]