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If:

$a = \sqrt{ b^2 - b }$

The problem I have is that for values of:

$0 < b < 1$

the result of:

$b^2 - b$

Is a negative number which gives rise to an error on Excel and my calculator.

I understand that negative numbers don't have square roots (I read it on Wikipedia at least), so how do I solve this for values of $b$ less than 1?

Thanks! :)

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    Good point Mark, I'll do that!2012-05-18

3 Answers 3

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Mathematicians have defined a new number, called $i$, such that $i^2=-1$ (it's not really that new). Commonly $i$ is called an imaginary number. If you're familiar with coordinate geometry like the Cartesian plane, the complex numbers are very similar. Every complex number has the form $a+bi$ for some $a,b\in\mathbb{R}$ so we can plot complex numbers (that is, numbers that have a real part and an imaginary part) as pairs (a,b) where we view the typical $x$-axis as the real part and the $y$-axis as the imaginary axis. If you have a number like $\sqrt{-64}$, you can simplify it by pulling out the $-1$ as an $i$. That is, $ \sqrt{-64}=i\sqrt{64}=\pm8i $ Complex numbers have lots of interesting properties. I recommend checking out the wikipedia page on complex numbers for more information.

Specifically to answer your question, if $b^2-b<0$, there are no solutions over the real numbers. You need to use complex numbers in order to find solutions.

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    @pglove don't forget to click the check mark if this answered your question!2012-05-20
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If $b^2-b<0$, then $b-b^2>0$ and $a = \pm i\sqrt{ b - b^2 },$ where $i$ is the imaginary unit, which by definition is the unique complex number that satisfies $i^2=-1\Leftrightarrow i=\pm\sqrt{-1}.$

The complex numbers are numbers of the form $a+bi$, where $a$ and $b$ are real numbers. They appear e.g. in the solution of a quadratic equation with negative discriminant, such as this one $x^2+x+1=0,$ whose solutions are $x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm\sqrt{-3}}{2}=\dfrac{-1\pm\sqrt{3}\ i}{2}.$

Example: For $b=1/2$, we have $b^2-b=1/4-1/2=-1/4$ and $a = \pm i\sqrt{ \frac{1}{2} - \frac{1}{4 }}=\pm i\sqrt{ \frac{1}{4} }=\pm \frac{1}{2}i .$ We could have computed as follows

$a = \sqrt{ \frac{1}{4 }-\frac{1}{2} }=\sqrt{ -\frac{1}{4} }=\sqrt{ -1}\sqrt{ \frac{1}{4} }=\sqrt{ -1}\frac{1}{2}=\pm i \frac{1}{2}=\pm \frac{1}{2}i .$

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    @pglove Glad to know.2012-05-18
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Aside from the use if i, one cannot take the square root of negative numbers.