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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 $\frac{2}{\beta}$

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

This Question has 1 answers.

We are given that $ \alpha $ and $ \beta $ are the zeroes of the polynomial:

$ 8x^2 - 5x - 1 = 0 $

### Step 1: Find Sum and Product of Zeroes
Using Vieta’s formulas:

- Sum of the zeroes: $ \alpha + \beta = \frac{-(-5)}{8} = \frac{5}{8} $
- Product of the zeroes: $ \alpha \beta = \frac{-1}{8} $

We need to form a quadratic polynomial whose zeroes are $ \frac{2}{\alpha} $ and $ \frac{2}{\beta} $.

### Step 2: Find the Sum and Product of New Zeroes
Sum of the new zeroes:

$ \frac{2}{\alpha} + \frac{2}{\beta} = 2 \left( \frac{\alpha + \beta}{\alpha \beta} \right) $

Substituting the values:

$ 2 \times \frac{\frac{5}{8}}{\frac{-1}{8}} = 2 \times \left( -\frac{5}{1} \right) = -10 $

Product of the new zeroes:

$ \frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha \beta} $

Substituting the value:

$ \frac{4}{\frac{-1}{8}} = 4 \times (-8) = -32 $

### Step 3: Form the Quadratic Polynomial
A quadratic polynomial with zeroes $ p $ and $ q $ is given by:

$ x^2 - (p+q)x + pq = 0 $

Substituting $ p+q = -10 $ and $ pq = -32 $:

$ x^2 + 10x - 32 = 0 $

### Final Answer:
The required quadratic polynomial is:

$ x^2 + 10x - 32 $

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