Multiplying surds with identities

Mul­ti­ply­ing out brack­ets with surds term by term always works, but three alge­braic iden­ti­ties make it faster and safer. They are the same iden­ti­ties you used in the chap­ters on alge­bra, now with surds in the place of let­ters. This les­son prac­tises them on squares of bino­mial surds, on con­ju­gate pairs, and on gen­eral prod­ucts.

The three iden­ti­ties#

$$\begin{aligned} (a + b)^{2} &= a^{2} + b^{2} + 2ab \\ (a - b)^{2} &= a^{2} + b^{2} - 2ab \\ (a + b)(a - b) &= a^{2} - b^{2} \end{aligned}$$

When $a$ or $b$ is a surd, remem­ber that $(\sqrt{b})^{2} = b$, and that a prod­uct such as $(\sqrt{7})(\sqrt{8})$ is $\sqrt{56}$, which should then be sim­pli­fied.

Squar­ing a bino­mial surd#

Exam­ple 1#

Find $(5 + \sqrt{6})^{2}$.

Use $(a + b)^{2} = a^{2} + b^{2} + 2ab$ with $a = 5$ and $b = \sqrt{6}$.

$$\begin{aligned} (5 + \sqrt{6})^{2} &= 5^{2} + (\sqrt{6})^{2} + 2(5)(\sqrt{6}) \\ &= 25 + 6 + 10\sqrt{6} \\ &= 31 + 10\sqrt{6} \end{aligned}$$

Exam­ple 2#

Find $(\sqrt{7} - \sqrt{8})^{2}$.

Use $(a - b)^{2} = a^{2} + b^{2} - 2ab$ with $a = \sqrt{7}$ and $b = \sqrt{8}$.

$$\begin{aligned} (\sqrt{7} - \sqrt{8})^{2} &= (\sqrt{7})^{2} + (\sqrt{8})^{2} - 2(\sqrt{7})(\sqrt{8}) \\ &= 7 + 8 - 2\sqrt{56} \\ &= 15 - 2\sqrt{56} \end{aligned}$$

This is cor­rect, but $\sqrt{56}$ is not in sim­plest form: $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$. So the final answer is

$$15 - 2 \times 2\sqrt{14} = 15 - 4\sqrt{14}$$

A gen­eral prod­uct#

Exam­ple 3#

Sim­plify $(3 + 2\sqrt{2})(5 - \sqrt{2})$.

There is no spe­cial iden­tity here, so mul­ti­ply term by term.

$$\begin{aligned} (3 + 2\sqrt{2})(5 - \sqrt{2}) &= 15 - 3\sqrt{2} + 10\sqrt{2} - 2(2) \\ &= 15 - 3\sqrt{2} + 10\sqrt{2} - 4 \\ &= 11 + 7\sqrt{2} \end{aligned}$$

The last prod­uct of the first line is $2\sqrt{2} \times (-\sqrt{2}) = -2 \times 2 = -4$.

Con­ju­gate pairs#

For a con­ju­gate pair, use $(a + b)(a - b) = a^{2} - b^{2}$. The answer is always ratio­nal.

Exam­ple 4#

Find $(4 - \sqrt{5})(4 + \sqrt{5})$.

$$\begin{aligned} (4 - \sqrt{5})(4 + \sqrt{5}) &= (4)^{2} - (\sqrt{5})^{2} \\ &= 16 - 5 \\ &= 11 \end{aligned}$$

Exam­ple 5#

Find $(7 + 4\sqrt{3})(7 - 4\sqrt{3})$.

$$\begin{aligned} (7 + 4\sqrt{3})(7 - 4\sqrt{3}) &= (7)^{2} - (4\sqrt{3})^{2} \\ &= 49 - 48 \\ &= 1 \end{aligned}$$

Here $(4\sqrt{3})^{2} = 16 \times 3 = 48$. Square the $4$ as well as the $\sqrt{3}$.

Two more prod­ucts#

Exam­ple 6#

Sim­plify $(5 + \sqrt{3})(7 + 4\sqrt{3})$.

$$\begin{aligned} (5 + \sqrt{3})(7 + 4\sqrt{3}) &= 35 + 20\sqrt{3} + 7\sqrt{3} + 4 \times 3 \\ &= 35 + 20\sqrt{3} + 7\sqrt{3} + 12 \\ &= 47 + 27\sqrt{3} \end{aligned}$$

You will meet this prod­uct again in the next les­son.

Exam­ple 7#

Sim­plify $(2\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})$.

$$\begin{aligned} (2\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) &= 2\sqrt{2} \cdot \sqrt{2} - 2\sqrt{2} \cdot \sqrt{3} + \sqrt{3} \cdot \sqrt{2} - \sqrt{3} \cdot \sqrt{3} \\ &= 4 - 2\sqrt{6} + \sqrt{6} - 3 \\ &= 1 - \sqrt{6} \end{aligned}$$

Prac­tice#

Sim­plify.

1. $(3 + \sqrt{2})^{2}$
2. $(\sqrt{5} - 2)^{2}$
3. $(2\sqrt{3} + \sqrt{2})^{2}$
4. $(\sqrt{6} - \sqrt{3})^{2}$
5. $(6 + \sqrt{7})(6 - \sqrt{7})$
6. $(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})$
7. $(4 + \sqrt{3})(2 - \sqrt{3})$
8. $(2 + 3\sqrt{5})(1 + \sqrt{5})$
9. $(\sqrt{3} + \sqrt{2})^{2} + (\sqrt{3} - \sqrt{2})^{2}$
10. $(1 + \sqrt{2})^{3}$

Answers#