If α and β are the zeroes of the polynomial \(x^2+5x+c\), and \( α-β=2\), then \(c=\)? a) 0 b) \(-\frac{21}{4}\) c) \(\frac{21}{4}\) d) 5

Answer: Option c. \(\frac{21}{4}\) Solution: Step 1: Use the concepts of relationship between roots and coefficients of quadratic polynomial. Given polynomial is \(x^2+5x+c\), by comparing it standard \quadratic polynomial \(ax^2+bx+c\), we get \(a=1, \quad b=5, \quad c=c\) \( \alpha+\beta= \frac{-b}{a}=\frac{-5}{1} = -5 \)            ....(1) \(\alpha \beta= \frac{c}{a}=\frac{c}{1} =c\)              ....(2) \(\alpha-\beta= 2\)     .....(3)    [given] Step 2: Find the value of α and β By adding (1) and (2), we get \(\alpha+\beta +\alpha-\beta= -5+2\) \(\Rightarrow 2\alpha = -3\) \(\Rightarrow \alpha = \frac{-3}{2}\) Put the value of \(α= \frac{-3}{2}\)  in (1), we get \(\frac{-3}{2}+\beta= -5\) \(\Rightarrow \beta= -5+\frac{3}{2}\) \(\Rightarrow \beta= \frac{-10+3}{2}\) \(\Rightarrow \beta= \frac{-7}{2}\) Step 3: Use the concept of product of roots of quadratic polynomial \(∵  \quad \alpha\times \beta = \frac{c}{a}\) \(\therefore  \quad \alpha\times \beta = \frac{c}{1}=c\) Substitute the value of \(\alpha\) and \(\beta\), we get \(c= \frac{-3}{2} \times \frac{-7}{2}\) \(\Rightarrow c=\frac{21}{4}\) Hence, the value of c is \(\frac{21}{4}\).