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We all know the solutions of

$$\sqrt{x}=x$$

are $x=0$ and $x=1$, which are easily checked. If I wanted to solve for the latter solution, I would proceed as follows:

\begin{eqnarray} \sqrt{x}&=&x\\ \left(\sqrt{x}\right)^2&=&\left(x\right)^2 \\ x &=& x^2 \\ 1 &=& x \end{eqnarray}

However, I am at a loss to solve the solution $x=0$ in a similar algebraic manner. Can anyone show me where to begin?

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    $\begingroup$ Be careful when you divide x from both sides. This is actually where you're getting rid of the x = 0 solution. You can instead move the xs to get 0 = x^2 - x = x(x-1), then you get 0 = x or 0 = x -1 -> x = 1 $\endgroup$
    – Jason Carr
    Sep 22, 2015 at 17:15
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    $\begingroup$ Set equal to $0$ and factor, once you have a polynomial. $\endgroup$
    – pjs36
    Sep 22, 2015 at 17:15
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    $\begingroup$ There's a step at the end where you divide by $x$, but that step is only valid if $x \neq 0$. $\endgroup$
    – littleO
    Sep 22, 2015 at 17:17

3 Answers 3

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Continuing from $x=x^2$:

$$x=x^2$$ $$x^2-x=0$$ $$x\cdot(x-1)=0$$ So, $x=0$ or $x=1$

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  • $\begingroup$ Wow, I feel retarded. Thanks! $\endgroup$ Sep 22, 2015 at 17:53
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Once you reach $x=x^2$, and you divide with $x$, you are already supposing, that $x \ne 0$, since if $x=0$, you can't divide. If you just stick with $x=x^2$, you will get $x_1=1, x_2=0 (x=x^2 \rightarrow 0=x^2-x=x(x-1))$

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When you go from the step $x = x^2$, you divide through the equation by $x$ which implicitly assumes $x\neq 0$, so you have to check if $x = 0$ is a solution as well. However you can also say

$$x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0 \implies x = 0 \text{ or } x = 1. $$

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