Show that $x-x^2=\frac{1}{3}$ has no real solutions.

Question

I have graphed this equation, which shows that there are no real solutions to this equation. How would I go about showing this without relying on a graph?

Answer 1

Convert it to the form $ax^2 +bx+c=0$. If $b^2 -4ac<0,$ then there are no real solutions. This is trivial when looking at the quadratic formula.

Answer 2

$$x^2-x+\frac13\gt x^2-x+\frac14=\left(x-\frac12\right)^2\ge0$$

Answer 3

I mean the easiest way is to use the discriminant (as mentioned in the first comment). But hey, why use the quadratic equation when you can just complete the square (i.e. derive the quadratic equation):

$$ x^2 - x +\frac{1}{3} = 0\\ \left(x - \frac{1}{2}\right)^2 + \frac{1}{3} + ? = 0 \\ x^2 - x + \frac{1}{4} + \frac{1}{3} + ? = 0 \leadsto ? = -\frac{1}{4}\\ \left(x - \frac{1}{2}\right)^2 + \frac{1}{3} - \frac{1}{4} = 0 \\ \left(x - \frac{1}{2}\right)^2 + \frac{1}{12}= 0 \\ \left(x - \frac{1}{2}\right)^2 = -\frac{1}{12} $$

A real value cannot be squared (positive or negative) to give a negative result, so this has no real solutions.

If we inspect $b^2 - 4ac$, we find:

$$ 1 - \frac{4}{3} = -\frac{1}{3} < 0 \leadsto\ \text{no real solutions} $$

...no surprise, if we do a little manipulation, we can see this value in our above equation:

$$ \left(2\left(x - \frac{1}{2}\right)\right)^2 = (2x - 1)^2 = 4\left(x - \frac{1}{2}\right)^2 $$

So we can use that and just multiply each side by 4

$$ (2x - 1)^2 = -\frac{1}{3} $$