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

  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) \(6x^2+8x+1\).