Question

Prove that 2x^2+x+8>0 for all real values of x.

Answer 1

$y = 2x^2 + x + 8$ is a polynomial of even degree, with a positive $x^2$ coefficient, meaning that
- it will have exactly one turning point
- that turning point will be a minimum

So, if the y-coordinate of the turning point is positive, then this polynomial will be positive for all real values of x.

At a turning point, the gradient of y will be equal to 0. The gradient of y is given by 
$\frac{\mathrm{d}y}{\mathrm{d}x} = 4x + 1$.

To find the turning point, set this equal to 0 and solve for x:
$4x + 1 = 0 \implies x = -\frac{1}{4}$.

Substituting this value into the equation gives
$y = 2(-\frac{1}{4})^2 + (-\frac{1}{4}) + 8 = \frac{1}{4} - \frac{1}{4} + 8 = 8 \ \textgreater \ 0$.

Since the minimum point of the equation is greater than 0, the equation will be greater than 0 for all real values of x.