Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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

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! :)

share|cite|improve this question

migrated from May 17 '12 at 5:53

This question came from our discussion, support, and feature requests site for people studying math at any level and professionals in related fields.

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. – Ben Millwood May 17 '12 at 13:19
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. – Mark Bennet May 17 '12 at 13:44
Good point Mark, I'll do that! – 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.

share|cite|improve this answer
Thank you, this is very clear and easy to understand for me :) – pglove May 18 '12 at 8:52
@pglove don't forget to click the check mark if this answered your question! – 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

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

share|cite|improve this answer
Thanks again, you helped me with my first question on the site too :) – pglove May 18 '12 at 8:53
@pglove Glad to know. – Américo Tavares May 18 '12 at 9:37

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.