If a | vedclass

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

Question

(A) Given the quadratic inequality $ax^2 - 2x + 4 > 0$ with $a < 0$.To find the roots of the corresponding equation $ax^2 - 2x + 4 = 0$,we use the qua

Vedclass · Accepted Answer

$\frac{1 + \sqrt{1 - 4a}}{a} > x > \frac{1 - \sqrt{1 - 4a}}{a}$

Vedclass · Answer

(A) Given the quadratic inequality $ax^2 - 2x + 4 > 0$ with $a To find the roots of the corresponding equation $ax^2 - 2x + 4 = 0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Here,$b = -2$ and $c = 4$,so $x = \frac{2 \pm \sqrt{(-2)^2 - 4(a)(4)}}{2a} = \frac{2 \pm \sqrt{4 - 16a}}{2a} = \frac{1 \pm \sqrt{1 - 4a}}{a}$.Since $a The inequality $ax^2 - 2x + 4 > 0$ holds between the two roots.Thus,the solution is $\frac{1 - \sqrt{1 - 4a}}{a} < x < \frac{1 + \sqrt{1 - 4a}}{a}$.

If $a < 0$,then the inequality $ax^2 - 2x + 4 > 0$ has the solution represented by:

Similar Questions

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 35x + 2 = 0$,then the value of $(2\alpha - 35)^3 \cdot (2\beta - 35)^3$ is equal to

The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in $[0, 1]$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module