Find the range of $x$ for which $\frac{3x-x^2-4}{-3x+x^2-4}<0$

Question

Find the range of values of $k$ for which $3x-x^2+k$ is negative for all real values of $x$. Hence, or otherwise, find the range of values of x for which $\frac{3x-x^2-4}{-3x+x^2-4} < 0$

I have gotten $k < \frac{-9}{4}$. Anyone knows how to do the second part where $\frac{3x-x^2-4}{-3x+x^2-4}<0$?

Answer 1

Note that as the leading coefficient in $-x^2+3x+k$ is negative, the equation will have negative values for all $k$ iff the determinant is negative. Hence:

$$D = 3^2 + 4k <0 \iff k < -\frac 94$$

So it seems you got that one right. Now note that $-4<k$ therefore by part one the numerator of the fraction is always negative. Now to find all solutions you need to find when the denominator is positive. We have that $x^2 - 3x - 4 = (x-4)(x+1)$. Therefore we have that the equation is true for $x \in (-\infty, -1) \cup (4,\infty)$

Answer 2

the coefficient of $x^2$ is negative so, the discriminant $\delta=9+4k$ must be negative. which gives $k\leq \frac{-9}{4}$.

for the second, the numerator discriminant is $-7 <0 $.

thus the numerator is always strictly negative (same signum than the coefficient of $x^2$). for denominator, the discriminant is $25$ and the roots are $ -1 $ and $4$. the denominator is strictly positive outside , let say in

$(-\infty,-1) U (4,\infty)$ which is your answer set.