I looked this up and seen something that was beyond my A-Level Maths course.
In class we are doing the discriminant and sketching quadratic graphs, so it is nothing advanced. My teacher completed the square to prove the quadratic:
$$ 2x^2 + 8x + 9 $$
is always positive for all real values of $x$. So it was:
$$2(x + 2)^2 + 1$$
And her notes have the $2$ multiplying the $2$ in the bracket saying "always positive", and pointing to the $1$ saying always positive. I just don't understand how that proves it, and what if it was negative?
Any help appreciated. 🙂
Best Answer
$(x + 2)^2 \ge 0$ because it is a square. So $2(x + 2)^2 + 1 \ge 1 > 0$.
As $2(x+2)^2 + 1 = 2x^2 + 8x + 9$, $2x^2 + 8x + 9 > 0$
In general, not all quadratics will be entirely positive or entirely negative but you can always convert $ax^2 + bx + c = a(x^2 + b/ax + b^2/4a^2) + c - b^2/a = a(x + b/2a)^2 + (c - b^2/a)$ The term squared will always be non-negative. If a and (c - b^2/a) are both positive or are both negative the quadratic will be either always positive or negative. If they are not both then the quadratic will be be positive for some values and negative for others.