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 $\cos^4 \theta + \alpha$ and $\sin^4 \theta + \alpha$ are the roots of the equation $x^2 + 2bx + b = 0$,and $\cos^2 \theta + \beta$ and $\sin^2 \theta + \beta$ are the roots of the equation $x^2 + 4x + 2 = 0$,then find the value of $b$.

The number of real roots of the equation ${e^{\sin x}} - {e^{-\sin x}} - 4 = 0$ is:

Let $r_1, r_2, r_3$ be the roots of the equation $x^3 - 2x^2 + 4x + 5074 = 0$. Then the value of $(r_1 + 2)(r_2 + 2)(r_3 + 2)$ is:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt{625}$
$II.$ $y = \sqrt{676}$

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the roots of the equation $cx^2 + bx + a = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module