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**Bacterius** First question : First, you know that $\displaystyle a> 0$. That means that if you sketch the graph for any permitted value of $\displaystyle p$, it will be a curved parabola open to the top. Therefore, there are only two possible situations : either it never touches the x-axis, therefore there are no real solutions and the equation is always positive, either it cuts the x-axis and there are solutions. But the equation cannot be all negative. Now, if $\displaystyle b^2 < 4ac$, that means that $\displaystyle b^2 - 4ac < 0$. Basically, it means that the discriminant is negative, thus the equation admits no real solution and therefore is positive for all real $\displaystyle x$ (you got to rewrite it properly though, I gave you the idea).

Second question : you need to find all values of $\displaystyle p$ for which the given quadratic has only positive values (that is, has no real solution). Think of your equation like this :

$\displaystyle ax^2 + bx + c$

Where :

$\displaystyle a = 4$

$\displaystyle b = 4p$

$\displaystyle c = -(3p^2 + 4p - 3)$

Get the discriminant of this equation :

$\displaystyle \Delta = b^2 - 4ac$

By substituting your values :

$\displaystyle \Delta = (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)]$

Now, what is the characteristic feature of a quadratic equation with no real solution (i.e. all values are positive, in our case) ? The discriminant is negative. Therefore you have :

$\displaystyle \Delta < 0$

$\displaystyle (4p)^2 - 4 \times 4 \times [-(3p^2 + 4p - 3)] < 0$

$\displaystyle 16p^2 + 16 \times (3p^2 + 4p - 3) < 0$

$\displaystyle 16p^2 + 48p^2 + 64p - 48 < 0$

$\displaystyle 64p^2 + 64p - 48 < 0$

Solve for $\displaystyle p$ and you will then have the range of $\displaystyle p$ when the quadratic equation has no real solution (always positive).

Does it help ? Ask if you still need help.

PS : why did you put a congruent sign ($\displaystyle \equiv$) on your function ?