The square root of a surd

Expres­sions like $\sqrt{5 + 2\sqrt{6}}$, a square root with a surd inside it, look awk­ward. But some of them sim­plify beau­ti­fully: $\sqrt{5 + 2\sqrt{6}}$ is exactly $\sqrt{3} + \sqrt{2}$. The trick is to recog­nise the num­ber under the root as a per­fect square, using the iden­tity $(a + b)^{2} = a^{2} + b^{2} + 2ab$. This last les­son of the chap­ter shows how.

The idea#

Square $\sqrt{3} + \sqrt{2}$:

$$(\sqrt{3} + \sqrt{2})^{2} = (\sqrt{3})^{2} + (\sqrt{2})^{2} + 2(\sqrt{3})(\sqrt{2}) = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$$

So $5 + 2\sqrt{6}$ is a per­fect square, and its square root is $\sqrt{3} + \sqrt{2}$. To run this back­wards on $\sqrt{p + 2\sqrt{q}}$, look for two num­bers that

• add to give $p$, and
• mul­ti­ply to give $q$.

If the num­bers are $m$ and $n$, then $p + 2\sqrt{q} = (\sqrt{m})^{2} + (\sqrt{n})^{2} + 2\sqrt{m}\sqrt{n} = (\sqrt{m} + \sqrt{n})^{2}$, and so

$$\sqrt{p + 2\sqrt{q}} = \sqrt{m} + \sqrt{n}, \qquad \sqrt{p - 2\sqrt{q}} = \sqrt{m} - \sqrt{n} \quad (m > n)$$

The method needs a $2$ in front of the inner root. If it is not there, make it appear first.

Exam­ple 1#

Find $\sqrt{5 + 2\sqrt{6}}$.

We need two num­bers that add to $5$ and mul­ti­ply to $6$: they are $3$ and $2$.

$$\begin{aligned} \sqrt{5 + 2\sqrt{6}} &= \sqrt{3 + 2 + 2\sqrt{3 \times 2}} \\ &= \sqrt{(\sqrt{3})^{2} + (\sqrt{2})^{2} + 2(\sqrt{3})(\sqrt{2})} \\ &= \sqrt{(\sqrt{3} + \sqrt{2})^{2}} \\ &= \sqrt{3} + \sqrt{2} \end{aligned}$$

The last step is just like $\sqrt{4^{2}} = 4$ or $\sqrt{9^{2}} = 9$.

A minus sign: keep the answer pos­i­tive#

A square root sign always means the pos­i­tive root. So $\sqrt{(-3)^{2}} = \sqrt{9} = 3$, not $-3$. In gen­eral, $\sqrt{t^{2}}$ is $t$ when $t$ is pos­i­tive and $-t$ when $t$ is neg­a­tive.

This mat­ters when the mid­dle sign is a minus. Both $(\sqrt{3} - \sqrt{2})^{2}$ and $(\sqrt{2} - \sqrt{3})^{2}$ equal $5 - 2\sqrt{6}$. But $\sqrt{2} - \sqrt{3}$ is neg­a­tive (about $-0.32$), so it can­not be the value of a square root. Always write the larger surd first.

Exam­ple 2#

Find $\sqrt{5 - 2\sqrt{6}}$.

$$\begin{aligned} \sqrt{5 - 2\sqrt{6}} &= \sqrt{3 + 2 - 2\sqrt{3 \times 2}} \\ &= \sqrt{(\sqrt{3})^{2} + (\sqrt{2})^{2} - 2(\sqrt{3})(\sqrt{2})} \\ &= \sqrt{(\sqrt{3} - \sqrt{2})^{2}} \\ &= \sqrt{3} - \sqrt{2} \end{aligned}$$

Since $\sqrt{3} > \sqrt{2}$, the answer $\sqrt{3} - \sqrt{2}$ is pos­i­tive, as it must be.

Mak­ing the 2 appear#

Exam­ple 3#

Find $\sqrt{7 - \sqrt{48}}$.

There is no $2$ in front of $\sqrt{48}$. But $48 = 4 \times 12$, so $\sqrt{48} = \sqrt{4} \times \sqrt{12} = 2\sqrt{12}$.

$$\begin{aligned} \sqrt{7 - \sqrt{48}} &= \sqrt{7 - 2\sqrt{12}} \\ &= \sqrt{4 + 3 - 2\sqrt{4 \times 3}} \\ &= \sqrt{(\sqrt{4})^{2} + (\sqrt{3})^{2} - 2(\sqrt{4})(\sqrt{3})} \\ &= \sqrt{(\sqrt{4} - \sqrt{3})^{2}} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3} \end{aligned}$$

The two num­bers adding to $7$ and mul­ti­ply­ing to $12$ are $4$ and $3$.

Exam­ple 4: mul­ti­ply­ing by 2 over 2#

Find $\sqrt{2 + \sqrt{3}}$.

Here $\sqrt{3}$ has a $1$ in front, not a $2$, and $3$ has no square fac­tor to bor­row from. Instead, mul­ti­ply and divide by $2$ under the root:

$$\displaystyle \begin{aligned} \sqrt{2 + \sqrt{3}} &= \sqrt{\frac{2(2 + \sqrt{3})}{2}} = \frac{\sqrt{4 + 2\sqrt{3}}}{\sqrt{2}} \\ &= \frac{\sqrt{3 + 1 + 2\sqrt{3 \times 1}}}{\sqrt{2}} \\ &= \frac{\sqrt{(\sqrt{3})^{2} + (1)^{2} + 2\sqrt{3}(1)}}{\sqrt{2}} \\ &= \frac{\sqrt{(\sqrt{3} + 1)^{2}}}{\sqrt{2}} = \frac{\sqrt{3} + 1}{\sqrt{2}} \end{aligned}$$

If you like, ratio­nalise the denom­i­na­tor too: $\displaystyle \frac{\sqrt{3} + 1}{\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{2}$.

Work­ing back­wards#

Exam­ple 5#

If $\sqrt{13 - a\sqrt{10}} = \sqrt{8} + \sqrt{5}$, find $a$.

Square both sides:

$$\begin{aligned} 13 - a\sqrt{10} &= (\sqrt{8} + \sqrt{5})^{2} \\ &= 8 + 5 + 2\sqrt{40} \\ &= 13 + 2 \times 2\sqrt{10} \\ &= 13 + 4\sqrt{10} \end{aligned}$$

Com­par­ing the surd parts, $-a = 4$, so $a = -4$. A neg­a­tive value is per­fectly cor­rect here: the left side is then $\sqrt{13 + 4\sqrt{10}}$.

Prac­tice#

1. Find $\sqrt{7 + 2\sqrt{10}}$.
2. Find $\sqrt{8 - 2\sqrt{15}}$.
3. Find $\sqrt{3 - 2\sqrt{2}}$.
4. Find $\sqrt{9 + 4\sqrt{2}}$.
5. Find $\sqrt{11 - \sqrt{120}}$.
6. Find $\sqrt{16 + 6\sqrt{7}}$.
7. Find $\sqrt{4 + \sqrt{15}}$, and ratio­nalise the denom­i­na­tor of your answer.
8. If $\sqrt{8 + b\sqrt{15}} = \sqrt{5} + \sqrt{3}$, find $b$.

Answers#