If the roots of {x^2} + x + a = 0 exceed a,th | vedclass

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(D) For the roots of the quadratic equation $f(x) = x^2 + x + a = 0$ to exceed a number $k = a$,the following conditions must be satisfied:$1$. $D = b

Vedclass · Accepted Answer

$a < -2$

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(D) For the roots of the quadratic equation $f(x) = x^2 + x + a = 0$ to exceed a number $k = a$,the following conditions must be satisfied:$1$. $D = b^2 - 4ac \ge 0$$2$. $f(k) > 0$$3$. $-\frac{b}{2a} > k$Applying these conditions for $k = a$:$1$. $1^2 - 4(1)(a) \ge 0 \implies 1 - 4a \ge 0 \implies a \le \frac{1}{4}$$2$. $f(a) = a^2 + a + a > 0 \implies a^2 + 2a > 0 \implies a(a + 2) > 0 \implies a  0$$3$. $-\frac{1}{2(1)} > a \implies a Taking the intersection of all three conditions:$a \le \frac{1}{4}$,$(a  0)$,and $a The intersection is $a < -2$.

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

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