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 \(\frac{21}{4}\).