vector spaces - square root of $1/2 + sqrt3/2?$

Playing with Maple, I noticed that it gives the square root of $c = 1+\frac{\sqrt3}{2}$ as equal to $a = \frac{1}{2}+\frac{\sqrt3}{2}$.

Indeed it checks out. But I got curious: how can I find that value, or more generally any square root of numbers of the form $x+y\sqrt{k}$?

I was able to do it the following way: the square of $z = x+y\sqrt{k}$ is also of the same form. Therefore, I can suppose there is a number of that form whose square is equal to $z$.

In my case, I want to find $(x,y)$ such that $(x+y\sqrt3)^2 = 1+\frac{\sqrt3}{2}$. I developed, which yields $(x^2+3y^2) + 2xy\sqrt3 = 1+\frac{\sqrt3}{2}$. Then I matched the coefficients of $1$ and of $\sqrt3$ on both sides, to get the system:

$x^2+3y^2 = 1$

$2xy = \frac{1}{2}$

Solving for $x$ and $y$, I got $x = ±\frac{1}{2}$ and $y = ±\frac{1}{2}$ (there is another pair of solutions that compute to the same number). QED.

Is my method correct? Is there any more efficient way? Is it possible to prove that a solution of the form $z = x+y\sqrt{k}$ always exist and if not, when?

Thanks.

Answer

Simpler than undetermined coefficients is the following rule I discovered as a teenager.

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Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

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Here $\:1+\sqrt{3}/2\:$ has norm $= 1/4.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1/2\ $ yields $\ 1/2+\sqrt{3}/2\:$

and this has $\rm\ \sqrt{trace}\: =\: 1,\ \ thus,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1/2+\sqrt{3}/2.$

Below is another example.

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Note $\:9-4\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 7\ $ yields $\ 2-4\sqrt{2}\:$

and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1-2\sqrt{2}.$

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See here for many more examples, and see this answer for general radical denesting algorithms.