# Explain why there's no solution to the equation $2x-2x^2 = 1$

How can you tell there's no solution to the equation $2x - 2x^2 = 1$.

The supporting information goes like this:

The diagram above shows the graph of $y = 7 + 2x - 2x^2$.

I tried to do this: $2 \cdot 1^2 = 2y$. therefore $y = 2-2 = 0$;

• What kinds of solutions are you allowing? If you allow complex solutions you will of course have solutions. Mar 16, 2013 at 21:51

The equation has no real solution because $2x - 2x^2 - 1 = -(x-1)^2-x^2 < 0$ for all $x\in\mathbb R$.

Alternatively (to lhf's more straightforward solution), you could just immediately check that the discriminant $b^2−4ac=4−8=−4<0.$ The discriminant is the term under the square root occuring in the quadratic formula. Since it is negative, the only solutions will be complex.

• This is also a great answer. It's a tough choice between your answer and lhf's. Mar 16, 2013 at 22:05

We want to solve $2x^2-2x+1=0$, presumably in the reals. Look at the upward-facing parabola $y=2x^2-2x+1$.

Complete the square. We have $x^2-x=(x-1/2)^2-1/4$. Thus our equation can be written as $$y=2(x-1/2)^2+1/2.$$ The vertex of the parabola is at $x=1/2$. There, $y=1/2$, so the parabola is always above the $x$-axis. Thus it has no point in common with the $x$-axis, and therefore our equation has no real solutions.

The arithmetic is a little easier if we look at the equivalent equation $4x^2-4x+2=0$, which can be rewritten as $(2x-1)^2+1=0$. But $(2x-1)^2$ is always $\ge 0$ if $x$ is real. So we can't have $(2x-1)^2+1=0$ for real $x$.

Another point of view: If you are looking at $2x^2-2x=-1$ in the integers, the left side is even and the right side is odd.

Let $$x=\frac12+t$$. Then $$t^2=-\frac14$$. So, $$t$$ and $$x$$ are not real.