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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 $y = 2-2 = 0$;

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

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

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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.

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  • $\begingroup$ This is also a great answer. It's a tough choice between your answer and lhf's. $\endgroup$
    – lrobb
    Mar 16, 2013 at 22:05
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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 $x$.

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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.

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

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