If the roots of the equation x^2+x+a=0 exceed | vedclass

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(B) Let $f(x) = x^2+x+a$. For the roots to exceed $a$,the following conditions must be satisfied:$1$. $D \geq 0$ $\Rightarrow 1-4a \geq 0$ $\Rightarro

Vedclass · Accepted Answer

$a < -2$

Vedclass · Answer

(B) Let $f(x) = x^2+x+a$. For the roots to exceed $a$,the following conditions must be satisfied:$1$. $D \geq 0$ $\Rightarrow 1-4a \geq 0$ $\Rightarrow a \leq \frac{1}{4}$.$2$. $f(a) > 0$ $\Rightarrow a^2+a+a > 0$ $\Rightarrow a^2+2a > 0$ $\Rightarrow a(a+2) > 0$ $\Rightarrow a \in (-\infty, -2) \cup (0, \infty)$.$3$. $-\frac{b}{2a} > a$ $\Rightarrow -\frac{1}{2} > a$ $\Rightarrow a Taking the intersection of all three conditions:$a \in (-\infty, \frac{1}{4}] \cap ((-\infty, -2) \cup (0, \infty)) \cap (-\infty, -\frac{1}{2})$$= (-\infty, -2)$.

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

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