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(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:

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