Values for p so inequality is right for every x. $-9< \frac{3x^2+px-6}{x^2-x+1} < 6 $

Question

$$-9< \frac{3x^2+px-6}{x^2-x+1} < 6 $$ I have done problems where I need to get the parameter so the inequality is right for real values of x. But what do I have to do to make it right for every value of x? What condition? I am not asking for anyone to throw me the answer of the whole thing. Just to clarify what the condition needs to be so I can finish on my own.

Answer 1

Since $x^2-x+1>0$, we obtain $$3x^2-(p+6)x+12>0$$ and $$12x^2+(p-9)x+3>0,$$ for which we need $$(p+6)^2<144$$ and $$(p-9)^2<144,$$ which gives $$-18<p<6$$ and $$-3<p<21,$$ which is $$-3<p<6.$$

Answer 2

hint

The denominator $x^2-x+1$ is always strictly positive.

multiplying by it, we get

$$-9x^2+9x-9 <3x^2+px-6 $$ and $$3x^2+px-6<6x^2-6x+6$$

thus

$$12x^2+(p-9)x+3 >0$$ and $$3x^2-(6+p)x+12>0$$

the discriminant should be negative.

$$(p-9)^2 <144$$ and $$(p+6)^2 <144$$

which give $$|p-9|<12 \iff -3 <p <21$$ and $$|p+6|<12 \iff -18 <p <6$$

finally, the answer is $$\color {red}{-3 <p <6}$$