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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$a < -2$

Vedclass · Answer

(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

Similar Questions

If one root of $x^2+px-q^2=0$, where $p$ and $q$ are real, is less than $2$ and the other is greater than $2$, then:

For what set of values of $K$ do both roots of the equation $4x^2 - 20Kx + (25K^2 + 15K - 66) = 0$ lie below $2$?

Let the set of all values of $p \in R$,for which both the roots of the equation $x^2-(p+2)x+(2p+9)=0$ are negative real numbers,be the interval $(\alpha, \beta]$. Then $\beta-2\alpha$ is equal to

For a real number $x$,if the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$,then:

The value of $p$ for which both the roots of the equation $4x^2 - 20px + (25p^2 + 15p - 66) = 0$ are less than $2$ lies in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module