Is there an easier way to solve the following equation?
$$x^2=2x-1$$
I think I know how to find $x$, using the quadratic formula:
I get
$$x^2-2x+1=0$$
then
$$x=\frac{2 \pm \sqrt{4-4})}2= \frac{2 \pm \sqrt{0}}2$$
but I don't know what $\sqrt{0}$ is. Is it $0$? If so, I would get $x=1$. Is that right?
The teacher said that $\sqrt{\phantom 0}$ is only for positives. Is $0$ positive?
$\endgroup$ 62 Answers
$\begingroup$The square root function, in the reals, $\sqrt a$ is defined for all $a\geq 0$: that means the square root of a real number $a$ is defined strictly for all $a$ greater than or equal to $0$.
So your root becomes $\dfrac {2\pm 0}{2} = 1$. This root has multiplicity of two; indeed, $$x^2 - 2x + 1 = (x-1)^2 = (x-1)(x-1)$$
$\endgroup$ 4 $\begingroup$it is equivalent to $$x^2-2x+1=0$$ and we get $$(x-1)^2=0$$
$\endgroup$ 1
