Question

Consider quadratic equation $a{x}^2+(2-a)x-2=0$, where $a\in R$.

If exactly one root is negative, then the range of $a^2+2a+5$ is

• A. $[4,\infty )$
• B. $[-2,\infty )$
• C. $(-\infty ,4]$
• D. $(5,\infty )$

Answer 1

The correct option is D $(5,\infty )$

$a{x}^2+(2-a)x-2=0$

For exactly one root negative means the roots will be of opposite signs i.e. $0$ lie between the roots

$af\left(0\right)<0$

$\Rightarrow a(-2)<0$

$\Rightarrow a>0$

$\Rightarrow a(a+2)>0$

$\Rightarrow a^2+2a>0$

$\Rightarrow a^2+2a+5>5$

So, Range is $(5,\infty )$