I have been asked this question:
Show that $x^2 + 2px + 2p^2$ is positive for all real values of $x$.
I've worked it out like so:
Discriminant = $(2p)^2 – (4\times 1\times(2p^2)) = 4p^2 – 8p^2$
I realise that the discriminant must be $\le0 $
No matter the value of $p$, $4p^2 – 8p^2$ will always be $\le0 $.
Also, by completing the square:
$x^2 + 2px + 2p^2 = (x+p)^2 +p^2$
Again, the $p^2$ value on the right will always be positive.
Therefore, no matter the value of p, the parabola will be positive for all values of $y$.
Am I correct?
I feel that there could be a more mathematical way of expressing this.
Best Answer
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.