### If sum of the zeroes = α+β = - 8 and product of zeroes = αβ = 6, then the polynomial whose zeroes are \(\frac{1}{α}\) and \(\frac{1}{β}\) is

**(a) \(6x^2+8x+1\)**

**(b) \(6x^2-8x-1\)**

**(c) \(6x^2-4x+6\)**

**(d) \(6x^2-8x+1\)**

**Answer:** Option (a) \(6x^2+8x+1\).

**Solution**:

**Step 1: Use the concept of formation of quadratic polynomial when roots are given**

The quadratic polynomial whose roots are \(\alpha\) and \(\beta\) is given by

\(x^2-(\alpha+\beta)x+\alpha\beta\)

So, the polynomial whose roots are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) is

\(x^2-(\frac{1}{\alpha}+\frac{1}{\beta})x+\frac{1}{\alpha} \times\frac{1}{\beta}\)

**Step 2: Simplify the coefficient of \(x\) and then use the given values of sum and product of zeroes**

\(x^2-(\frac{\alpha+\beta}{\alpha \beta})x+\frac{1}{\alpha \beta}\)

\(x^2-(\frac{-8}{6})x+\frac{1}{6}\) **[ ∵ \(\alpha+\beta=-8, \alpha \beta=6\)]**

\(\Rightarrow x^2+\frac{8}{6}x+\frac{1}{6}\)

\(\Rightarrow \frac{6x^2+8x+1}{6}\)

\(\Rightarrow \frac{1}{6} \times(6x^2+8x+1)\)

Here, \(\frac{1}{6} \) is the constant.

So, the required polynomial is \(6x^2+8x+1\).

**Hence, the correct option is option (a) \(6x^2+8x+1\).**