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I am a little confused with the following calculation of a radical.

$$\left(-\frac {\sqrt{-r}}{2} \right)^2=-\frac{r}{4}$$

$$\left(\frac {\sqrt{-r}}{2} \right)^2=-\frac{r}{4}$$

I would have thought both of these solutions would be,

$$\frac {r}{4}$$

So why are they both negative? Just looking for the details of this calculation

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    yes that is my question – 2017-02-12

4 Answers 4

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Assuming we're dealing with real numbers here, $r$ is forced to be negative. So let's plug in a number ($r = -4$) and examine more closely what is going on:

$$\left(-\frac {\sqrt{-(-4)}}{2} \right)^2=\left(-\frac{\sqrt 4}{2}\right)^2 = (-1)^2 = 1 = \dfrac {-4}{-4} = -\dfrac {r}{4}$$

So we see that none of the rules were broken:

$1)$ We only took a square root of a positive number

$2)$ Squaring yielded a positive number

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Hints:

We know that $(a)^2 = (-a)^2 = a^2$ and that $\sqrt{-a} = \sqrt{-1}\times \sqrt{a} = \sqrt{a}i$. Now, use the fact $i^2=-1$ where $i$ is the imaginary unit. Thus, $(\sqrt{-a})^2 = (\sqrt{a})^2 \times (i)^2 = (a) \times (-1) = -a$.

Hope it helps.

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Note that $$\left( \sqrt{x} \right)^2=x$$ And $$\left(\frac{x}{y} \right)^2=\frac{x^2}{y^2}$$ So $$\left(\frac {\sqrt{-r}}{2} \right)^2=\frac{\left( \sqrt{-r} \right)^2}{2^2}=-\frac{r}{4}$$

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    why would you take the negative sign out like that? shouldn't it remain in the brackets? – 2017-02-12
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    @fr14 Yeah, fixed. – 2017-02-12
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    okay so you would square the negative sign making it positive then remove the radical sign without changing the value of the number inside the radical making it a negative then? – 2017-02-12
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    @fr14 Yes. We do that. We don't change the value inside the radical. – 2017-02-12
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    thanks for your solution btw it was very helpful, I will accept the answer when the timer allows – 2017-02-12
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2:
$$\left(\frac {\sqrt{-r}}{2} \right)^2=\left(\frac {\sqrt{-1}\times\sqrt{r}}{2} \right)^2=\left(\frac {i\sqrt{r}}{2} \right)^2$$

and $i^2 = -1$.

1: $$\left(-\frac {\sqrt{-r}}{2} \right)^2=\left(i\frac {\sqrt{r}}{2} \right)^2 \times (-1)^2$$