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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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    $\begingroup$ 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. $\endgroup$ Commented May 17, 2012 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. $\endgroup$ Commented May 17, 2012 at 13:44
  • $\begingroup$ Good point Mark, I'll do that! $\endgroup$
    – pglove
    Commented May 18, 2012 at 8:55
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    $\begingroup$ Considering the context, I assume that you want to stay in the real numbers. And here, there simply is no square root of a negative number simply because a square is always non-negative. So, the expression is not defined within the real numbers, hence cannot be calculated. Additionally, "solved" should be rather used in the case of equalities or inequalities. An expression is "evaluated" or "calculated". $\endgroup$
    – Peter
    Commented Aug 8, 2020 at 9:00

4 Answers 4

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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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  • $\begingroup$ Thank you, this is very clear and easy to understand for me :) $\endgroup$
    – pglove
    Commented May 18, 2012 at 8:52
  • $\begingroup$ @pglove don't forget to click the check mark if this answered your question! $\endgroup$ Commented May 20, 2012 at 17:01
  • $\begingroup$ Sorry for the EXTREMELY outdated comment, but how is $\sqrt{64}$ equal to $\pm8$ here? Isn't $\sqrt{ }$ the principle square root? $\endgroup$
    – AdamLee123
    Commented Oct 18, 2023 at 15:35
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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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  • $\begingroup$ Thanks again, you helped me with my first question on the site too :) $\endgroup$
    – pglove
    Commented May 18, 2012 at 8:53
  • $\begingroup$ @pglove Glad to know. $\endgroup$ Commented May 18, 2012 at 9:37
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Aside from the use if i, one cannot take the square root of negative numbers.

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Svenkat's and Google answers: "one cannot take the square root of negative numbers" "Negative numbers don't have real square roots" "you can't take the square root of a negative number"

It is very difficult to break 16th century scholastic dogma, e.g. imaginary numbers, used in taking the square root of negative numbers. Nobody wants to discuss or in other words - taboo. If we are consistent, we should accept to take square roots only from positive numbers. The result of multiplication of the same numbers is always a positive number. If we take the square root of a negative number, we should place the negative sign before the radical sign/radix -√. Details: https://globid.eu/

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  • $\begingroup$ Negativism and negative votes are widely practiced. Please vote and comment to understand your vote. $\endgroup$
    – Georgi
    Commented Aug 12, 2020 at 5:40

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