Simply:
Suppose $p\ne 0$, we have:
$
x^2+2px+2p^2=x^2+2px+p^2+p^2=(x+p)^2+p^2
$
so, as a sum of two squares is always positive.
This is true for $x,p \in \mathbb{R}$ since we know that the square of any rela number is positive and the sum of two positive numbers is positive.
Hint: Consider any $p$ for which the roots are real. The quadratic will have a positive and negative root if the constant term, $p+5$, is negative. If the constant term is positive, the two roots have the same sign, and we can use the fact that their sum is $-2(p-1)$.
Full solution:
By my first statement: if $p < -5$, then the roots have opposite sign, so we have at least one positive root.
If $p = -5$, then we have the quadratic $x^2 - 12x$, which has a positive root.
If $-5 < p \leq -1$, then we have two roots of the same sign. Since the sum of the roots is $-2(p-1) > 0$, we may conclude that both roots are positive.
If $p \geq 4$, then both roots have the same sign. Since the sum of the roots is $-2(p-1) < 0$, we may conclude that both roots are negative.
So, we have at least one positive root exactly when $p \leq -1$.
Quicker solution:
If $p \leq -1$, the sum of the roots is $-2(p - 1) > 0$. Conclude that we have a positive root.
If $p \geq 4$, then both roots have the same sign since $p+5 > 0$. Since the sum of the roots is $-2(p-1) < 0$, we may conclude that both roots are negative.
Best Answer
$x^2+x+1$
$=(x^2+2x+1)-x$
$=(x+1)^2-x$
Now,$(x+1)$ $>$ $x$.
So,$(x+1)^2$ $>$ $x$
So,$(x+1)^2$ $-$ $x$ is positive.