Multiplying and dividing real numbers

Adding roots needed like roots. Mul­ti­ply­ing is friend­lier: any two roots can be mul­ti­plied. In this les­son you will mul­ti­ply real num­bers that con­tain roots, see how a root times itself becomes a plain num­ber, and divide by roots. Two rules from the chap­ter on square roots do almost all the work.

The two rules#

$$\sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \qquad \sqrt{a} \times \sqrt{a} = \sqrt{a \times a} = a$$

Here $a$ and $b$ are $0$ or more. The first rule joins two roots into one. The sec­ond says a root times itself gives back the num­ber under the root.

Exam­ple 1#

$$\sqrt{3} \times \sqrt{3} = 3 \qquad \sqrt{3} \times \sqrt{6} = \sqrt{3 \times 6} = \sqrt{18}$$

Mul­ti­ply­ing num­bers with roots in front#

When each num­ber has a plain part and a root, mul­ti­ply the plain num­bers together and the roots together. You may change the order of a mul­ti­pli­ca­tion, so this is always allowed.

Exam­ple 2#

Find $2\sqrt{3} \times 5\sqrt{2}$.

$$\begin{aligned} 2\sqrt{3} \times 5\sqrt{2} &= 2 \times 5 \times \sqrt{3} \times \sqrt{2} \\ &= 10 \times \sqrt{6} \\ &= 10\sqrt{6} \end{aligned}$$

Exam­ple 3#

Find $2\sqrt{5} \times 5\sqrt{7}$.

The $2$ joins the $5$, and the $\sqrt{5}$ joins the $\sqrt{7}$.

$$2\sqrt{5} \times 5\sqrt{7} = 10\sqrt{35}$$

Exam­ple 4: the same root twice#

Find $2\sqrt{3} \times 5\sqrt{3}$ and $5\sqrt{6} \times 3\sqrt{6}$.

When the roots are the same, they become a plain num­ber.

$$\begin{aligned} 2\sqrt{3} \times 5\sqrt{3} &= 10 \times 3 = 30 \\ 5\sqrt{6} \times 3\sqrt{6} &= 15 \times 6 = 90 \end{aligned}$$

Exam­ple 5: a neg­a­tive num­ber#

Find $5\sqrt{5} \times (-3\sqrt{2})$.

A plus times a minus gives a minus.

$$5\sqrt{5} \times (-3\sqrt{2}) = -15\sqrt{10}$$

Exam­ple 6#

Find $2\sqrt{8} \times (-3\sqrt{2})$.

$$\begin{aligned} 2\sqrt{8} \times (-3\sqrt{2}) &= -6 \times \sqrt{8 \times 2} \\ &= -6 \times \sqrt{16} \\ &= -6 \times 4 \\ &= -24 \end{aligned}$$

A com­mon slip is to write $-96$, by mul­ti­ply­ing $6 \times 16$. But the $16$ is still under the root, and $\sqrt{16} = 4$.

Exam­ple 7#

Find $3\sqrt{7} \times (-2\sqrt{343})$.

$$\begin{aligned} 3\sqrt{7} \times (-2\sqrt{343}) &= -6 \times \sqrt{7 \times 343} \\ &= -6 \times \sqrt{2401} \\ &= -6 \times 49 \\ &= -294 \end{aligned}$$

To find $\sqrt{2401}$, use long divi­sion. The pairs are $24 \mid 01$. First, $4 \times 4 = 16$ leaves $24 - 16 = 8$. Bring down $01$ to make $801$. Dou­ble $4$ to get $8$, and $89 \times 9 = 801$ leaves noth­ing. So $\sqrt{2401} = 49$.

Remem­ber. Plain num­bers mul­ti­ply with plain num­bers, and roots with roots. A root times itself is the plain num­ber under it.

A num­ber out­side a bracket#

A num­ber out­side a bracket mul­ti­plies every part inside it.

$$a(b + c) = ab + ac$$

The grid shows $6 \times (2 + 5)$. Six rows of seven squares, split into a part $2$ wide and a part $5$ wide. Both ways give $42$: $6 \times 7 = 42$, and $6 \times 2 + 6 \times 5 = 12 + 30 = 42$. This is called the dis­trib­u­tive prop­erty.

Exam­ple 8#

Mul­ti­ply out $\sqrt{2}(3 + \sqrt{2})$.

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

The $3\sqrt{2}$ and the $2$ are not alike, so they stay apart.

Divid­ing real num­bers#

Three rules help with divi­sion.

$$\displaystyle \frac{a}{\sqrt{a}} = \sqrt{a} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \qquad \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$

The first one works because $a = \sqrt{a} \times \sqrt{a}$, and one $\sqrt{a}$ can­cels:

$$\displaystyle \frac{a}{\sqrt{a}} = \frac{\sqrt{a} \times \sqrt{a}}{\sqrt{a}} = \sqrt{a}$$

Exam­ple 9#

Find $\displaystyle \frac{5}{\sqrt{5}}$.

$$\displaystyle \frac{5}{\sqrt{5}} = \frac{\sqrt{5} \times \sqrt{5}}{\sqrt{5}} = \sqrt{5}$$

Exam­ple 10#

Find $\displaystyle \frac{7 \times \sqrt{49}}{\sqrt{7}}$.

First, $\sqrt{49} = 7$, so the top is $7 \times 7$. Now write one of the 7s as $\sqrt{7} \times \sqrt{7}$.

$$\displaystyle \frac{7 \times \sqrt{49}}{\sqrt{7}} = \frac{7 \times \sqrt{7} \times \sqrt{7}}{\sqrt{7}} = 7\sqrt{7}$$

A com­mon slip is to answer $\sqrt{49}$. Check: $7\sqrt{7} \times \sqrt{7} = 7 \times 7 = 49$, which is the top. So $7\sqrt{7}$ is right.

Exam­ple 11#

Find $\displaystyle \frac{\sqrt{8}}{2\sqrt{2}}$.

Split $8$ into $4 \times 2$, because $4$ is a per­fect square.

$$\displaystyle \begin{aligned} \frac{\sqrt{8}}{2\sqrt{2}} &= \frac{\sqrt{4 \times 2}}{2\sqrt{2}} \\ &= \frac{\sqrt{4} \times \sqrt{2}}{2\sqrt{2}} \\ &= \frac{2\sqrt{2}}{2\sqrt{2}} \\ &= 1 \end{aligned}$$

A com­mon slip is to answer $2$. That is $\displaystyle \frac{\sqrt{8}}{\sqrt{2}}$; but the bot­tom here is $2\sqrt{2}$.

Exam­ple 12: mov­ing a root off the bot­tom#

Find $\displaystyle \frac{8}{\sqrt{2}}$ with no root on the bot­tom.

Mul­ti­ply the top and the bot­tom by $\sqrt{2}$. That is mul­ti­ply­ing by $\displaystyle \frac{\sqrt{2}}{\sqrt{2}} = 1$, so the value does not change.

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

Never divide by zero#

You may divide by any real num­ber except $0$. A num­ber like $\displaystyle \frac{2}{0}$ has no mean­ing.

Com­mon mis­takes#

• Mul­ti­ply­ing the num­ber under the root by the num­ber in front. In $-6 \times \sqrt{16}$, first find $\sqrt{16} = 4$.
• Leav­ing $\sqrt{a} \times \sqrt{a}$ as $\sqrt{a^2}$. Fin­ish it: it is $a$.
• Mul­ti­ply­ing only the first part of a bracket. $\sqrt{2}(3 + \sqrt{2})$ needs both parts mul­ti­plied.
• Divid­ing by $0$.

Prac­tice#

1. Find $5\sqrt{3} \times (-12\sqrt{3})$.
2. Find $5\sqrt{2} \times (-8\sqrt{2})$.
3. Find $\sqrt{5} \times (-3\sqrt{5})$.
4. Find $1787\sqrt{2} \times 3\sqrt{2}$.
5. Find $\sqrt{5} \times \sqrt{23}$.
6. Find $7\sqrt{6} \times 8\sqrt{3}$, leav­ing the answer as a num­ber times one root.
7. Find $\sqrt{12} \times \sqrt{3}$.
8. Find $\displaystyle \frac{\sqrt{8}}{\sqrt{2}}$.
9. Find $\displaystyle \frac{2}{\sqrt{2}}$.
10. Write $\displaystyle \frac{6}{\sqrt{3}}$ with no root on the bot­tom.
11. Mul­ti­ply out $\sqrt{2}(5 + \sqrt{2})$.

Answers#