Properties of square roots

Square roots fol­low some sim­ple rules. A few of them let you mul­ti­ply and divide roots quickly. Oth­ers are warn­ings: things that look true but are not. Know­ing both kinds will save you from the most com­mon mis­takes with square roots. In this les­son $a$ and $b$ stand for num­bers that are $0$ or more, so their square roots make sense.

Rule 1: mul­ti­ply­ing roots#

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

Two square roots mul­ti­plied together make one square root of the prod­uct.

Exam­ple 1#

$$\sqrt{2} \times \sqrt{8} = \sqrt{2 \times 8} = \sqrt{16} = 4$$

Exam­ple 2#

$$\sqrt{45} \times \sqrt{20} = \sqrt{45 \times 20} = \sqrt{900} = \sqrt{30 \times 30} = 30$$

Nei­ther $\sqrt{45}$ nor $\sqrt{20}$ is a whole num­ber, but their prod­uct is. Rule 1 makes this easy.

Rule 2: a root times itself#

$$\left(\sqrt{a}\right)^2 = \sqrt{a} \times \sqrt{a} = \sqrt{a^2} = a$$

A square root mul­ti­plied by itself gives back the num­ber under the root. That is just what a square root means.

$$\left(\sqrt{2}\right)^2 = \sqrt{2} \times \sqrt{2} = 2 \qquad \left(\sqrt{3}\right)^2 = \sqrt{3} \times \sqrt{3} = 3$$

Rule 3: divid­ing roots#

$$\displaystyle \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \qquad (b \text{ not } 0)$$

Exam­ple 3#

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

Exam­ple 4#

Find $\displaystyle \frac{\sqrt{243}}{\sqrt{363}}$.

Put both under one root and can­cel the com­mon fac­tor $3$.

$$\displaystyle \frac{\sqrt{243}}{\sqrt{363}} = \sqrt{\frac{243}{363}} = \sqrt{\frac{3 \times 81}{3 \times 121}} = \sqrt{\frac{81}{121}} = \frac{9}{11}$$

The last step takes the root of the top and the bot­tom sep­a­rately: $\sqrt{81} = 9$ and $\sqrt{121} = 11$.

Rule 4: a num­ber over its own root#

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

Why? By Rule 2, $a = \sqrt{a} \times \sqrt{a}$. So

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

because one $\sqrt{a}$ can­cels.

$$\displaystyle \frac{2}{\sqrt{2}} = \sqrt{2} \qquad \frac{3}{\sqrt{3}} = \sqrt{3} \qquad \frac{5}{\sqrt{5}} = \sqrt{5}$$

Warn­ings: what you must not do#

The next rules say what is not true. The sign $\neq$ means "is not equal to".

Rule 5: you can­not add under one root#

$$\sqrt{a} + \sqrt{b} \neq \sqrt{a + b}$$

Test it with $a = 25$ and $b = 16$.

Left side: $\sqrt{25} + \sqrt{16} = 5 + 4 = 9$.

Right side: $\sqrt{25 + 16} = \sqrt{41}$, which is not even a whole num­ber.

Since $9 = \sqrt{81}$, the left side is $\sqrt{81}$ and the right side is $\sqrt{41}$. They are not equal.

Rule 6: you can­not take away under one root#

$$\sqrt{a} - \sqrt{b} \neq \sqrt{a - b}$$

With $a = 25$ and $b = 16$: the left side is $5 - 4 = 1$, but the right side is $\sqrt{9} = 3$.

Rule 7: the root of a sum of squares#

$$\sqrt{a^2 + b^2} \neq a + b$$

Test it with $a = 3$ and $b = 4$.

$$\text{Left side} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

$$\text{Right side} = 3 + 4 = 7$$

$5$ is not $7$, so the two sides are not equal.

Rule 8: the root of a square of a sum#

$$\sqrt{(a + b)^2} = a + b$$

This one is true, because the square root undoes the square. With $a = 3$ and $b = 4$:

$$\sqrt{(3 + 4)^2} = \sqrt{7^2} = 7 \qquad \text{and} \qquad 3 + 4 = 7$$

Com­pare Rules 7 and 8 care­fully. $3^2 + 4^2 = 25$, but $(3 + 4)^2 = 49$. Squar­ing first and then adding is not the same as adding first and then squar­ing.

Remem­ber. $a^2 + b^2$ is not the same as $(a + b)^2$.

Rule 9: the root of a dif­fer­ence of squares#

$$\sqrt{a^2 - b^2} \neq a - b$$

With $a = 5$ and $b = 4$: the left side is $\sqrt{25 - 16} = \sqrt{9} = 3$. The right side is $5 - 4 = 1$.

Rule 10: the root of a square of a dif­fer­ence#

$$\sqrt{(a - b)^2} = a - b \qquad (a \text{ bigger than } b)$$

With $a = 5$ and $b = 4$: $\sqrt{(5 - 4)^2} = \sqrt{1^2} = 1$, and $5 - 4 = 1$.

Using the rules together#

Exam­ple 5#

Find $\sqrt{12} \times \sqrt{3}$.

$$\sqrt{12} \times \sqrt{3} = \sqrt{36} = 6$$

Exam­ple 6#

Find $\displaystyle \frac{\sqrt{98}}{\sqrt{2}}$.

$$\displaystyle \frac{\sqrt{98}}{\sqrt{2}} = \sqrt{\frac{98}{2}} = \sqrt{49} = 7$$

Exam­ple 7#

Find $\sqrt{6^2 + 8^2}$.

Do not write $6 + 8$. Square first, then add, then take the root.

$$\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Com­mon mis­takes#

• Writ­ing $\sqrt{9} + \sqrt{16} = \sqrt{25}$. The left side is $3 + 4 = 7$, and $\sqrt{25} = 5$.
• Writ­ing $\sqrt{3^2 + 4^2} = 3 + 4$. It is $\sqrt{25} = 5$.
• For­get­ting that $\sqrt{a} \times \sqrt{a}$ is just $a$, not $\sqrt{a^2}$ left unfin­ished.

Prac­tice#

1. Find $\sqrt{3} \times \sqrt{27}$.
2. Find $\sqrt{18} \times \sqrt{8}$.
3. Find $\sqrt{5} \times \sqrt{20}$.
4. Find $\left(\sqrt{7}\right)^2$.
5. Find $\displaystyle \frac{\sqrt{75}}{\sqrt{3}}$.
6. Find $\displaystyle \frac{\sqrt{128}}{\sqrt{2}}$.
7. Write $\displaystyle \frac{7}{\sqrt{7}}$ in a sim­pler way.
8. Is $\sqrt{9} + \sqrt{16} = \sqrt{25}$ true? Show why.
9. Find $\sqrt{13^2 - 12^2}$.
10. Find $\sqrt{(6 + 8)^2}$.

Answers#