# $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational

If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$

Then what is the value of $a$? The answer is $-1$.

$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$

$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,

$$\sqrt{u^2 - \sqrt{3}u} = a + b\sqrt{3}$$

$$u^2 - \sqrt{3}u = u(u - \sqrt{3})$$

$$a + b\sqrt{3} = \sqrt{u}\sqrt{u - \sqrt{3}}$$

What should I do?

• Compute $(a+b\sqrt{3})^2$. Compare coefficients. Feb 16, 2015 at 13:30
• $$4 - 2\sqrt{3} = a^2 + 2\sqrt{3}ab + 3b^2$$ mmm.... coeff $a^2=0$ but this makes $a = 0$? Feb 16, 2015 at 13:33
• No, $4 - 2\sqrt{3} = (a^2 + 3b^2) + (2ab)\sqrt{3}$. Which parts are rational, which irrational? Feb 16, 2015 at 13:34
• I see, $let b = 1$ and let $a=-1$ gives the proper answer. Feb 16, 2015 at 13:35
• Note that \begin{align*} \sqrt{4 - 2\sqrt{3}} & = \sqrt{3 - 2\sqrt{3} + 1}\\ & = \sqrt{\sqrt{3}^2 - 2 \cdot 1 \cdot \sqrt{3} + 1^2}\\ & = \sqrt{(\sqrt{3} - 1)^2}\\ & = \sqrt{3} - 1\end{align*} Feb 16, 2015 at 15:34

Such square roots can be computed by a Simple Denesting Rule:

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

which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{4}\, =\, 2.\ \ \rm \color{brown}{Dividing\ it\ out}\$ of the above yields $\ \ 1-\sqrt 3$

Finally, since the result is negative, we need to negate it, which yields $\ \sqrt 3 - 1$

Remark $\$ Many more worked examples are in prior posts on this denesting rule.

• (+1), what is the trace? Feb 16, 2015 at 17:13
• @Lebes Follow the first link for the definitions. Feb 16, 2015 at 17:27
• (+1),This rule is pretty amazing. Much simpler than my approach. I'll have to remember this. Feb 16, 2015 at 18:30

We can start with $\sqrt{4-2\sqrt{3}} = a + b\sqrt{3}$ and square both sides. This gives us

$$4-2\sqrt{3} = a^2 + 2ab\sqrt{3} + 3b^2$$

Because $a$ and $b$ are said to be rational, we know that the only term which will be a multiple of $\sqrt{3}$ is $2ab\sqrt{3}$ From this, we can split our system into two equations:

$$2ab = -2$$

and

$$a^2 + 3b^2 = 4$$

Let's use the first equation to sub $b = -1/a$. This gives us

$$a^2 + 3/a^2 = 4 \Rightarrow a^4 -4a^2 + 3 = 0$$

The quadratic formula then gives us $a = 1$ and $a = -1$, which would make $b = -1$ and $b = 1$, respectively. Because $\sqrt{3} > 1$ and the square root is always defined as positive, $\sqrt{4-2\sqrt{3}} = 1 - \sqrt{3}$ is an extraneous solution. Therefor, the only solution is $a = -1$ and $b = 1$.

Using formula $$\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}$$ you will get $$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt{12}}=\sqrt{\dfrac{4+\sqrt{16-12}}{2}}-\sqrt{\dfrac{4-\sqrt{16-12}}{2}}=-1+\sqrt3$$ So, $a=-1$ and $b=1$.