Question

$x^2+px+5=0$ has mean of first $n$ natural number as one root, then value of $p$ is

• A. $-\frac{n^2+2n+21}{2(n+1)}$
• B. $-\frac{n^2+2n+21}{n-1}$
• C. $-\frac{n^2+2n+6}{2(n+1)}$
• D. $-\frac{n^2+2n+6}{n-1}$

Answer 1

Answer: (A)

$-\frac{n^2+2n+21}{2(n+1)}$

Given equation is $x^2+px+5=0$

Given that $\frac{1+2+3+...+n}{n}=\frac{n+1}{2}$ is one root of given expression

Let the another root be $a$

We know that product of roots i.e. $a\times \frac{n+1}{2}=5$

$\Rightarrow a=\frac{10}{n+1}$

Sum of roots is equal to $\frac{n+1}{2}+\frac{10}{n+1}=-p$

$\Rightarrow p=-\left(\frac{n^2+2n+21}{2(n+1)}\right)$