Show that $x^2-8x+17>0$ for all real values of x Could someone please help me with this question. I thought I had to use the discriminant, which I found to be -4 but now I'm not sure if i'm just meant to factorise, as the discriminant would show there are no real solutions. 
Thank you :)

If the discriminant is negative then there are no values where the expression is equal to zero and as values can't "jump" from positive to negative without "passing through" zero then all values must be either all positive or all negative.
And as for $x=0$ this takes a positive value then all values are positive.
In general, if the determinate is $D = b^2 - 4ac$ and $a\ne 0$ then 
$ax^2 + bx + c = $
$a(x + 2\frac {b}{2a}x + \frac {b^2}{4a^2}) + c - \frac {b^2}{4a} =$
$a(x + \frac b{2a})^2 + \frac {4ac - b^2}{4a} = $
$a(x+\frac b{2a})^2 - \frac D{4a}$.
Now $(x + \frac b{2a})^2 \ge 0$ because it is a square.
So if $a > 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 - \frac D{4a} \ge -\frac D{4a}$.
If $D < 0$ then this is always a positive value.
If $a < 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 -\frac D{4a} \le -\frac D{4a}$.  If $D < 0$ then this is always a negative value. (Note: $D$ and $a$ are both negative so the "cancel each other out").
(Note: if $D = b^2 - 4ac < 0$ then $ac > \frac {b^2}4 \ge 0$.  So $a$ and $c$ must either both be positive of both be negative.)
... 
But it's easier to just complete the square:
$x^2 -8x + 17 = $
$(x^2 - 8x) + 17 = $
$(x^2 - 8x + 16) +17 - 16 = $
$(x-4)^2 + 1$.
And $(x-4)^2 \ge 0$ so $(x-4)^2 + 1 \ge 1$.
