Simplifying square roots

The answer $\sqrt{18}$ is cor­rect, but it is not as sim­ple as it could be. Hid­den inside $18$ is the per­fect square $9$, and $\sqrt{9} = 3$ can come out of the root. In this les­son you will learn to take per­fect squares out of a root. This tidies answers, and it also shows when two roots that look dif­fer­ent are really like roots that can be added.

The idea#

The rule $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ works in both direc­tions. So if the num­ber under the root has a per­fect square as a fac­tor, split it off.

$$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$$

The $4$ came out of the root as $2$. The $2$ stayed inside, because it is not a per­fect square.

Roots that become a num­ber times root 2#

Look at this fam­ily. Each num­ber is a per­fect square times $2$.

$$\begin{aligned} \sqrt{8} &= \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2} \\ \sqrt{18} &= \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} \\ \sqrt{32} &= \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} \\ \sqrt{50} &= \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \\ \sqrt{72} &= \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \\ \sqrt{98} &= \sqrt{49 \times 2} = \sqrt{49} \times \sqrt{2} = 7\sqrt{2} \end{aligned}$$

Roots that become a num­ber times root 3#

$$\begin{aligned} \sqrt{12} &= \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \\ \sqrt{27} &= \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3} \\ \sqrt{48} &= \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \\ \sqrt{75} &= \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3} \end{aligned}$$

And one with $5$: $\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$.

How to do it#

1. Look for the biggest per­fect square that divides the num­ber. Try $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$.
2. Write the num­ber as that square times what is left.
3. Take the square root of the square out­side. Leave the rest inside.

If you choose a square that is not the biggest, you will have to do it again. For $\sqrt{72}$, using $4$ gives $\sqrt{72} = 2\sqrt{18}$, and $\sqrt{18}$ can still be tidied to $3\sqrt{2}$. That gives $2 \times 3\sqrt{2} = 6\sqrt{2}$ in the end. Using $36$ straight away is quicker.

Exam­ple 1#

Sim­plify $\sqrt{28}$, $\sqrt{40}$ and $\sqrt{125}$.

$$\begin{aligned} \sqrt{28} &= \sqrt{4 \times 7} = 2\sqrt{7} \\ \sqrt{40} &= \sqrt{4 \times 10} = 2\sqrt{10} \\ \sqrt{125} &= \sqrt{25 \times 5} = 5\sqrt{5} \end{aligned}$$

For $125$, divide by $5$ again and again: $125$, $25$, $5$. So $125 = 25 \times 5$.

Exam­ple 2: using prime fac­tors#

Sim­plify $\sqrt{180}$.

If the square is hard to spot, break the num­ber into primes and pair them, as you did for square roots.

$$\begin{aligned} \sqrt{180} &= \sqrt{(2 \times 2) \times (3 \times 3) \times 5} \\ &= 2 \times 3 \times \sqrt{5} \\ &= 6\sqrt{5} \end{aligned}$$

Each pair comes out as one num­ber. The $5$ has no part­ner, so it stays inside.

Remem­ber. Split off the biggest per­fect square and take its root out­side. What is left stays under the root.

Adding roots that look dif­fer­ent#

$\sqrt{12}$ and $\sqrt{27}$ look unlike. But after tidy­ing, both are counts of $\sqrt{3}$.

Exam­ple 3#

Find $\sqrt{12} + \sqrt{27}$.

$$\begin{aligned} \sqrt{12} + \sqrt{27} &= 2\sqrt{3} + 3\sqrt{3} \\ &= 5\sqrt{3} \end{aligned}$$

Exam­ple 4#

Find $\sqrt{8} + \sqrt{32} + \sqrt{72}$.

$$\begin{aligned} \sqrt{8} + \sqrt{32} + \sqrt{72} &= 2\sqrt{2} + 4\sqrt{2} + 6\sqrt{2} \\ &= 12\sqrt{2} \end{aligned}$$

Exam­ple 5#

Find $5201\sqrt{12} - 13012\sqrt{12} + 780128\sqrt{12}$.

These are like roots already. Num­bers added: $5201 + 780128 = 785329$. Then $785329 - 13012 = 772317$.

$$5201\sqrt{12} - 13012\sqrt{12} + 780128\sqrt{12} = 772317\sqrt{12}$$

Now tidy the root: $\sqrt{12} = 2\sqrt{3}$.

$$772317\sqrt{12} = 772317 \times 2\sqrt{3} = 1544634\sqrt{3}$$

Tidy­ing after mul­ti­ply­ing#

Exam­ple 6#

Find $\sqrt{3} \times \sqrt{6}$ in its sim­plest form.

$$\sqrt{3} \times \sqrt{6} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$$

Exam­ple 7#

Find $7\sqrt{6} \times 8\sqrt{3}$ in its sim­plest form.

$$\begin{aligned} 7\sqrt{6} \times 8\sqrt{3} &= 56\sqrt{18} \\ &= 56 \times 3\sqrt{2} \\ &= 168\sqrt{2} \end{aligned}$$

Com­mon mis­takes#

• Tak­ing out a num­ber that is not a per­fect square. $\sqrt{12}$ is not $3\sqrt{4}$; the $4$ is the square, so $\sqrt{12} = 2\sqrt{3}$.
• For­get­ting to take the root of the square. $\sqrt{18} = \sqrt{9 \times 2}$ becomes $3\sqrt{2}$, not $9\sqrt{2}$.
• Stop­ping too early, as in $\sqrt{72} = 2\sqrt{18}$. Check whether the root left inside can be tidied again.
• Adding before tidy­ing. $\sqrt{12} + \sqrt{27}$ is $5\sqrt{3}$, not $\sqrt{39}$.

Prac­tice#

1. Sim­plify $\sqrt{45}$.
2. Sim­plify $\sqrt{63}$.
3. Sim­plify $\sqrt{80}$.
4. Sim­plify $\sqrt{200}$.
5. Sim­plify $\sqrt{108}$.
6. Sim­plify $\sqrt{162}$.
7. Find $\sqrt{12} + \sqrt{75}$.
8. Find $\sqrt{98} - \sqrt{50}$.
9. Find $\sqrt{125} - \sqrt{20}$.
10. Find $\sqrt{2} \times \sqrt{6}$ in its sim­plest form.
11. Find $3\sqrt{8} \times \sqrt{3}$ in its sim­plest form.

Answers#