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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?

amWhy
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Jacob
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2 Answers2

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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, $$x^2 - 2x + 1 = (x-1)^2 = (x-1)(x-1)$$

amWhy
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  • Thanks a lot. So my way was good also? You and Dr. Graubner seem to see the solution in a much simpler way by saying (x-1)^2. Why is this relevent? Why true? – Jacob Dec 12 '14 at 20:43
  • Your solution was fine. If you can see how to factor a quadratic easily, do so. But if not, you've always got the quadratic formula to use. The roots of your quadratic are those x-values that make the quadratic$ = 0$. Writing the function as $(x-1)^2$ shows us that $(x-1)^2 = 0 $ if and only if $(x-1) = 0,\iff x = 1$. – amWhy Dec 12 '14 at 20:44
  • Ok I see thank you. i think factor is next in my course i will ask teacher – Jacob Dec 12 '14 at 20:46
  • $x = a$ is a root, if and only if $(x - a)$ is a factor of the quadratic. (Note that when $x = a$, $(x - a) = a-a = 0$. By the zero product rule, if any one factor in a product of factors is zero, then the entire product is zero. – amWhy Dec 13 '14 at 15:21
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it is equivalent to $$x^2-2x+1=0$$ and we get $$(x-1)^2=0$$

Dr. Sonnhard Graubner
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