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

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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  • $\begingroup$ There's a reasonable question here as to why you think there is a solution. If $a = \sqrt{b^2 - b}$ then $a^2 = b^2 - b$, so when $b^2 - b$ is negative you are really saying "I know that the square of a number is always positive but what if it isn't?". As pointed out in answers, there are contexts where in fact the square of a number is not positive, but I think you should also just accept that sometimes an equation can have no solutions, or many solutions, and you can't just go ahead and solve for whatever you want. $\endgroup$ – Ben Millwood May 17 '12 at 13:19
  • $\begingroup$ Just to note that if you want to work with such things in Excel, you can use an IF(condition,value if true,value if false) function to bypass the error. $\endgroup$ – Mark Bennet May 17 '12 at 13:44
  • $\begingroup$ Good point Mark, I'll do that! $\endgroup$ – pglove May 18 '12 at 8:55

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.

  • $\begingroup$ Thank you, this is very clear and easy to understand for me :) $\endgroup$ – pglove May 18 '12 at 8:52
  • $\begingroup$ @pglove don't forget to click the check mark if this answered your question! $\endgroup$ – Milosz Wielondek May 20 '12 at 17:01

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

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

  • $\begingroup$ Thanks again, you helped me with my first question on the site too :) $\endgroup$ – pglove May 18 '12 at 8:53
  • $\begingroup$ @pglove Glad to know. $\endgroup$ – Américo Tavares May 18 '12 at 9:37

Aside from the use if i, one cannot take the square root of negative numbers.


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