# How do you prove that sqrt(4+2sqrt(3)) = sqrt(3)+1 ?

Nov 20, 2016

One way to show that the left hand side is equal to the right hand side is to show that their quotient is equal to $1$. Beginning with the quotient, we have

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1}$

To help us evaluate this, let's first rationalize the denominator

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right)}{\left(\sqrt{3} + 1\right) \times \left(\sqrt{3} - 1\right)}$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right)}{{\left(\sqrt{3}\right)}^{2} - {1}^{2}}$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right)}{3 - 1}$

$= \frac{\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right)}{2}$

As the quotient must be equal to $1$ if the given expressions are equal, we now need to show that the numerator is equal to $2$.

$\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right) = \sqrt{4 + 2 \sqrt{3}} \times \sqrt{{\left(\sqrt{3} - 1\right)}^{2}}$

(Note that the above step is justified because $\sqrt{3} - 1 > 0$. If $x \ge 0$, then $x = \sqrt{{x}^{2}}$. If $x < 0$, then $x = - \sqrt{{x}^{2}}$)

$= \sqrt{\left(4 + 2 \sqrt{3}\right) {\left(\sqrt{3} - 1\right)}^{2}}$

$= \sqrt{\left(4 + 2 \sqrt{3}\right) \left(3 - 2 \sqrt{3} + 1\right)}$

=sqrt((4+2sqrt(3))(4-2sqrt(3))

$= \sqrt{{4}^{2} - {\left(2 \sqrt{3}\right)}^{2}}$

$= \sqrt{16 - 12}$

(As when rationalizing the denominator, we make use of the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$)

$= \sqrt{4}$

$= 2$

Now that we have shown that the numerator has the desired property, we can solve the rest of the problem quite simply.

$\frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} = \frac{\sqrt{4 + 2 \sqrt{3}} \times \left(\sqrt{3} - 1\right)}{\left(\sqrt{3} + 1\right) \times \left(\sqrt{3} - 1\right)} = \frac{2}{2} = 1$

$\implies \frac{\sqrt{4 + 2 \sqrt{3}}}{\sqrt{3} + 1} \times \left(\sqrt{3} + 1\right) = 1 \times \left(\sqrt{3} + 1\right)$

$\therefore \sqrt{4 + 2 \sqrt{3}} = \sqrt{3} + 1$

Nov 20, 2016

See below.

#### Explanation:

This expression has the structure

$\sqrt{a + b \sqrt{3}} = c \sqrt{3} + d$ so squaring both sides

$a + b \sqrt{3} = 3 {c}^{2} + 2 \sqrt{3} c d + {d}^{2}$ pairing terms

$\left\{\begin{matrix}a - 3 {c}^{2} - {d}^{2} = 0 \\ b - 2 c d = 0\end{matrix}\right.$

Solving for $c , d$ we have

$c = \pm \frac{\sqrt{a \pm \sqrt{{a}^{2} - 3 {b}^{2}}}}{\sqrt{6}}$
$d = \pm \frac{\sqrt{\frac{3}{2}} b}{\sqrt{a - \sqrt{{a}^{2} \pm 3 {b}^{2}}}}$

If $a = 4 , b = 2$ we have the possibilities

$\left(\begin{matrix}c = - \frac{1}{\sqrt{3}} & d = - \sqrt{3} \\ c = \frac{1}{\sqrt{3}} & d = \sqrt{3} \\ c = - 1 & d = - 1 \\ c = 1 & d = 1\end{matrix}\right)$

Nov 20, 2016

See description...

#### Explanation:

Note that:

${\left(\sqrt{3} + 1\right)}^{2} = {\left(\sqrt{3}\right)}^{2} + 2 \left(\sqrt{3}\right) + 1$

$\textcolor{w h i t e}{{\left(\sqrt{3} + 1\right)}^{2}} = 3 + 2 \sqrt{3} + 1$

$\textcolor{w h i t e}{{\left(\sqrt{3} + 1\right)}^{2}} = 4 + 2 \sqrt{3}$

Since $\sqrt{3} + 1 > 0$, we can take the positive square root of both ends to get:

$\sqrt{3} + 1 = \sqrt{4 + 2 \sqrt{3}}$