# Finding the square root $\sqrt{4-2\sqrt{3}}$

How do you solve this:

$$X^2=4-2\sqrt{3}$$

I don't really know how to solve for $X$. I initially got $X=\sqrt{4-2\sqrt{3}}$ but I don't know how to simplify it.

$\sqrt{4-2\sqrt{3}}$ is a solution. So too is $-\sqrt{4-2\sqrt{3}}$

These particular expressions can be simplified (others not):

• the $2\sqrt{3}$ might suggest something of the form $(y-\sqrt{3})^2=4-2\sqrt{3}$ where $y=x+\sqrt{3}$

• so solve $y^2-2y\sqrt{3}-1+2\sqrt{3} = 0$

• giving $y=1$ or $y=2\sqrt{3}-1$,

• leading to $x=1-\sqrt{3}$ or $x=\sqrt{3}-1$

• Hmm. How is Y=X+√3? And why is it not applicable to other questions – Ashalley Samuel May 27 '17 at 9:24
• @AshalleySamuel: I suspect there is not a general rule: for example I doubt $\sqrt{4-\sqrt{3}}$ or $\sqrt{5-2\sqrt{3}}$ can be simplified. But in the case of $\sqrt{4-2\sqrt{3}}$ I guessed that something like completing the square might be interesting and it was: slightly more generally $\sqrt{x+1-2\sqrt{x}} = \sqrt{x}-1$ for $x \ge 1$ – Henry May 27 '17 at 12:55

$\surd$ simplifies if what's inside is a square.

Since $(x+y)^2=x^2+2xy+y^2$ we will search for $\begin{cases} x^2+y^2=4 \\ 2xy=-2\sqrt{3}\end{cases}\iff\begin{cases} x^2y^2+y^4=4y^2 \\ xy=-\sqrt{3}\end{cases}$

The quadratic equation $\big[y^4-4y^2+3=0\big]$ gives $y^2=1$ or $3$ and this leads to $x+y=\pm(1-\sqrt{3})$

If you want to explore further some methodology, I invite you to read this interesting paper :

Jeffrey-Rich : Simplifying square roots of square roots by denesting

Applying theorem $\mathbf{(4.12)}$ : $\sqrt{\vphantom{\big(}X\pm Y}=\sqrt{\frac 12X+\frac 12\sqrt{X^2-Y^2}}\pm\sqrt{\frac 12X-\frac 12\sqrt{X^2-Y^2}}$

With $X=4$ and $Y=2\sqrt{3}$

Gives $\sqrt{\vphantom{\big)}4-2\sqrt{3}}=\sqrt{2+\frac 12\sqrt{4}}-\sqrt{2-\frac 12\sqrt{4}}=\sqrt{3}-\sqrt{1}=\sqrt{3}-1$

Notice that $$4=3+1=\sqrt3^2+1^2$$

and you have a remarkable identity.

Let $\,a=\sqrt{4-2\sqrt{3}}\,$, $\,b=\sqrt{4+2\sqrt{3}}\,$ and note that $b \gt a \gt 0\,$, then:

$$\require{cancel} \begin{cases} a^2+b^2 = 4-\cancel{2\sqrt{3}} + 4+\cancel{2\sqrt{3}} = 8 \\ ab = \sqrt{(4-2\sqrt{3})(4+2\sqrt{3})} = \sqrt{16 - 2 \cdot 3} = 2 \end{cases}$$

It follows that:

\begin{cases} \begin{align} a+b &= \sqrt{a^2+b^2+2ab}=\sqrt{12}=2 \sqrt{3} \\ b-a &= \sqrt{a^2+b^2-2ab}=\sqrt{4} = 2 \end{align} \end{cases}

Therefore $a = \frac{1}{2}\big((a+b)-(b-a)\big)=\frac{1}{2}\left(2 \sqrt{3} - 2\right)=\sqrt{3}-1\,$.

Note that $4-2\sqrt{3}$ is simply $\left(\sqrt{3}-1\right)^2$. Thus, $\sqrt{4-2\sqrt{3}}$ can be simplify as: $$\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1$$