$x^2-px+p^2-4<0$ for at least one $x<0$ Find the set of all values of $p$, for which $x^2-px+p^2-4<0$ for at least one $x<0$.
My work:
Let $x=k<0$ be a root of this inequality. So our inequality becomes $$k^2-pk+p^2-4<0$$
Here $k^2$ is $>0$, $-pk$ is $>0$ So for the inequality to hold $p^2-4$ has to be $<0$.
After solving I got the range as, $p\in(-2,2)$. Now consider a quadratic equation $$f(y)=y^2-py+p^2-4=0$$ where $f(x)<0$ So here the $\triangle\ge0$ So $$p^2-4p^2+16\ge0$$ or $$p\in\left(-\infty,\frac{-4}{\sqrt{3}}\right]\cup\left[\frac{4}{\sqrt{3}},\infty\right)$$ Doing the intersection we get $$p\in\phi$$ But the answer is given as $p\in\left(\frac{-4}{\sqrt{3}},2\right)$ And I'm not sure about the correctness of the answer
But if it is correct, Where did I go wrong$?$
 A: *

*Your function is positive for very large $x$, so you never have the case where the quadratic is always negative. That means you require to have two real roots. If you have only one root, the value of the quadratic is never negative (always positive, except at the root, where it is $0$).

*$\Delta=p^2-4p^2+16\gt 0$means $p\in\left(-\frac4{\sqrt 3},\frac4{\sqrt 3}\right)$ (you got the wrong interval)

*If you want the quadratic to be negative for at least one $x<0$, it means you have the position of the vertex of the parabola at a value $\le 0$, or the value of the quadratic evaluated at $0$ is negative. The position of the vertex is at $p/2$. So one possibility is $p\le 0$. The other is $0^2-0p+p^2-4<0$, which means $p\in(-2,2)$. Since $2<4/\sqrt 3$, it means that $$p\in \left(-\frac4{\sqrt 3},2\right)$$
A: 
Find the set of all values of $p$, for which $x^2-px+p^2-4<0$ for at least one $x<0$.

In other words, we want the solution set of $$\text{for some negative } x, \:\:x^2-px+p^2-4 \text{ is negative}.\tag1$$
Since $x^2-px+p^2-4$ is an upwards-facing the parabola, we require

*

*two $x$-intercepts, and

*either a symmetry-axis with a nonpositive $x$-value or a negative $y$-intercept,

that is,

*

*$(-p)^2-4(1)(p^2-4)>0,\;$ and

*$-\frac{-p}{2(1)}\le0\quad$ or $\quad p^2-4<0,$
which is equivalent to

*

*$-\frac4{\sqrt3}<p<\frac4{\sqrt3},\;$ and

*$p\le0\quad$ or $\quad -2<p<2,$
which is equivalent to

*

*$-\frac4{\sqrt3}<p<\frac4{\sqrt3},\;$ and

*$p<2,$
which is equivalent to $$p\in\left(-\frac4{\sqrt3},2\right).$$
