Square root of $\sqrt{700} + \sqrt{280}$

Question

Is there a way to express $\sqrt{\sqrt{700}+\sqrt{280}}$ under only one radical sign like what is done to the example below? $$\sqrt{9+4\sqrt5}=\sqrt{4+5+2\sqrt{20}}=\sqrt{(\sqrt4+\sqrt5)^2}=2+\sqrt5$$

** I have done not much of a generalization but when I try to express $\sqrt{A+B}$ in the form of $\,R\,(a+\sqrt{b}\,)$ where $A$>$B$, $R$ is an adjustable ratio, and with $A$, $B$, $a$ and $b$ not being necessarily rational or irrational, I have arrived at this : $$a=\sqrt{b}\;\left(k\,\pm\sqrt{k^2-1}\,\right)$$ where $k = A/B$

In this equation I can plug in an arbitrary $b$ and find $a$ and after that make up a suitable value of $R$ to equal things up (I wonder if there exists such phrasal verb? ).

The problem with this is that, in $\sqrt{\sqrt{700}+\sqrt{280}}$ where $k=\sqrt{2.5}$, the R seems to always become another radicalsign-inside-radicalsign irrational number with a $k$ value of exactly $\sqrt{2.5}$ again regardless of what the value of $a$ and $b$ is. Is it that there isn't really a methodology to do it or that there is indeed a way and I have been missing things out? Thanks in advance.

Answer 1

1. Let me resolve your first problem in a general way.

Suppose there are two numbers $x$ and $y$ such that

$$ \sqrt{9 + 4\sqrt5} = \sqrt{(x + y\sqrt5)^2} = x + y\sqrt5\quad (1)$$

Square both sides:

$$ 9 + 4\sqrt5 = x^2 + 5y^2 + 2xy\sqrt5 $$ This equation can be satisfied by the following choice: $$ \begin{align} x^2 + 5y^2 &= 9 \quad (2) \\ 2xy &= 4 \quad (3)\end{align} $$

From $(3) \Rightarrow y = 2/x$, plug into $(2)$: $$ x^4 -9x^2 + 20 = (x^2 - 4)(x^2 - 5) = 0 $$ $$ x^2 = 4, 5 \Rightarrow x = 2,\sqrt5$$ where we have rejected the negative values of x as required by $(1)$. Therefore, there are two solutions: $$(x,y) = (2,1), (\sqrt5, \frac{2}{\sqrt5}). \quad (4)$$

Plugging both solutions to $(1)$, we obtain the same result:

$$ \sqrt{9 + 4\sqrt5} = 2 + \sqrt5 $$

2. Now consider the problem $A = \sqrt{\sqrt{700} + \sqrt{280}}$.

We will use the same method above, noting the following irreducible radical form:

$$\begin{align} \sqrt{700} &= 10\sqrt7, \\ \sqrt{280} &= 2\sqrt{2\cdot5\cdot7} = 2\sqrt{10}\sqrt7 \end{align} $$

Thus, $$ A = \sqrt{ \sqrt7(10 + 2 \sqrt{10}) } = \sqrt[4]7\sqrt{10 + 2 \sqrt{10}} . \quad (5)$$

Suppose there are two numbers $m$ and $n$ that satisfy:

$$ \sqrt{ 10 + 2 \sqrt{10} } = \sqrt{ (m + n\sqrt{10})^2 } = m + n\sqrt{10}. \quad(6) $$

Squaring both sides: $$ m^2 + 10 n^2 = 10, \quad (7)$$ $$ 2mn = 2 \Rightarrow n = 1/m. \quad (8)$$ It follows: $$ \begin{align} (6) \Rightarrow m^4 -10m^2 + 10 &= 0 \\ (m^2 -5)^2 -15 &= 0\\ m^2 &= 5 \pm \sqrt{15} \end{align} $$ Therefore, the solutions are as follows, after rejecting negative solutions as required by (6): $$ \begin{align} m &= \sqrt{5 \pm \sqrt{15}}, \\ n &= \frac{1}{\sqrt{5 \pm \sqrt{15}}} \\ &=\frac{\sqrt{5 \pm \sqrt{15}}}{5 \pm \sqrt{15}} \\ &= \frac{(5\mp\sqrt{15}) \sqrt{5 \pm \sqrt{15}} }{10}. \end{align}$$

When plugging the values of $m$ and $n$ to (6), we can see that the right hand side of (6) cannot be simplified to $a + b\sqrt{10}$ where $a$ and $b$ are rational numbers.

Thus, we can conclude that it is impossible to write

$$ A = \sqrt{ \sqrt7(10 + 2 \sqrt{10}) } = \sqrt[4]7 (a + b \sqrt{10}), $$

where $a$ and $b$ are rational numbers.

As the final note, let us remark that (6) can also be recast as

$$ \sqrt{ 10 + 2 \sqrt{10} } = \sqrt{ (x\sqrt2 + y\sqrt5)^2 } = x\sqrt2 + y\sqrt5.$$

Using the same procedure to solve for $x$ and $y$, we can prove that this form does not lead to a solution where $x$ and $y$ are rational numbers.