Question

lf both the roots of the equation $x^2+x+a=0$ exceed $a$, then

• A. $2<a<3$
• B. $a>3$
• C. $-3<a<3$
• D. $a<-2$

Answer 1

The correct option is D $a<-2$

For both roots to exist,

$D>0$

$\Rightarrow 1-4a>0$

$a<\frac{1}{4}$

$\Rightarrow a\in (-\infty ,\frac{1}{4})$

For both roots to exceed $a$, $f\left(a\right)>0$

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

$a(a+2)>0$

$a\in (-\infty ,-2)\bigcup (0,-\infty )$

$\therefore a\in (-\infty ,-2)$ for roots to exists and exceed $a$.

$\Rightarrow a<-2$