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 $a < 0$.First,find the roots of the corresponding quadratic equation $ax^2 - 2x + 4 = 0$ u

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 First,find the roots of the corresponding quadratic equation $ax^2 - 2x + 4 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Here,$x = \frac{2 \pm \sqrt{(-2)^2 - 4(a)(4)}}{2a} = \frac{2 \pm \sqrt{4 - 16a}}{2a} = \frac{1 \pm \sqrt{1 - 4a}}{a}$.Since $a The inequality $ax^2 - 2x + 4 > 0$ holds between the two roots.Let $\alpha = \frac{1 + \sqrt{1 - 4a}}{a}$ and $\beta = \frac{1 - \sqrt{1 - 4a}}{a}$.Since $a 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 the inequality $kx^2 - 2x + k \geq 0$ holds good for at least one real $x$,then the complete set of values of $k$ is

The integral value of $k$ for which $x^2 - 2(4k - 1)x + 15k^2 - 2k - 7 > 0$ for all $x \in R$ is

Let $f(x) = x^2 + 4x + 1$. Then

The set of all values of $x$ for which the inequalities $x^2-7x+10 \geq 0$ and $2x+3-x^2 > 0$ hold simultaneously is

The set of all values of $x$ which satisfy both the inequations $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$ simultaneously is

Vedclass Test Series

Exam Paper Generator

Online Exam Module