How do I prove a quadratic is always positive or negative for x?

Question

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. :)

Answer 1

$(x + 2)^2 \ge 0$ 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 $ax^2 + bx + c = a(x^2 + b x/a + b^2/(4a^2)) + c - b^2/(4a) = a(x + b/(2a))^2 + c - b^2/(4a)$. The term squared will always be non-negative. If $a$ and $c - b^2/(4a)$ 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.

Answer 2

whatever thing squared will always be positive. think about it, however small of a negative number x can be, after adding 2 to it, will still be vastly negative. however, when any negative number is squared it becomes negative - the negative of a negative is always positive.

Answer 3

The square of a real number is always positive (doesn't have to be strictly positive, but still positive).

$2(x+2)^2 + 1$ is always positive because it is the sum of two positive numbers. $2(x+2)^2$ is the product of two positive numbers so it is positive. 1 is positive because it is a strictly positive number.

$2x^2+8x+9 = 2(x+2)^2 + 1 \geq 1 > 0$ so $2x^2+8x+9$ is not only always positive, but always strictly positive for all real $x$.