Rationalising factors

In the last les­son, $(2 + \sqrt{3})(2 - \sqrt{3})$ came out as $1$, a plain ratio­nal num­ber. Pairs of surds that mul­ti­ply to give a ratio­nal num­ber have a name, and they are the key to the main skill of this chap­ter: remov­ing surds from the bot­tom of a frac­tion. This les­son defines the ratio­nal­is­ing fac­tor and shows how to find one for each com­mon kind of surd.

The def­i­n­i­tion#

If the prod­uct of two surds is a ratio­nal num­ber, then each of them is called a ratio­nal­is­ing fac­tor (R.F.) of the other.

For exam­ple, $\sqrt{2} \times \sqrt{2} = 2$, which is ratio­nal. So $\sqrt{2}$ is a ratio­nal­is­ing fac­tor of $\sqrt{2}$.

A surd has many ratio­nal­is­ing fac­tors. Since $\sqrt{2} \times 2\sqrt{2} = 4$ and $\sqrt{2} \times 3\sqrt{2} = 6$, the surds $2\sqrt{2}$, $3\sqrt{2}$ and so on are ratio­nal­is­ing fac­tors of $\sqrt{2}$ too. Any ratio­nal mul­ti­ple of a ratio­nal­is­ing fac­tor is again a ratio­nal­is­ing fac­tor. In prac­tice we use the sim­plest one.

Ratio­nal­is­ing fac­tors of the com­mon forms#

The last two rows come from the iden­tity $(x + y)(x - y) = x^{2} - y^{2}$. Squar­ing a square root removes it, so the prod­uct has no surd left:

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

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

So for a bino­mial surd, the con­ju­gate is a ratio­nal­is­ing fac­tor.

Exam­ple 1#

Find a ratio­nal­is­ing fac­tor of $3 + \sqrt{2}$.

Its con­ju­gate is $3 - \sqrt{2}$, and

$$(3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$$

which is ratio­nal. So $3 - \sqrt{2}$ is a ratio­nal­is­ing fac­tor.

Exam­ple 2#

Find a ratio­nal­is­ing fac­tor of $\sqrt{8} - \sqrt{7}$.

$$(\sqrt{8} - \sqrt{7})(\sqrt{8} + \sqrt{7}) = 8 - 7 = 1$$

So $\sqrt{8} + \sqrt{7}$ is a ratio­nal­is­ing fac­tor.

Exam­ple 3#

Find a ratio­nal­is­ing fac­tor of $7 + 4\sqrt{3}$.

Try the con­ju­gate, $7 - 4\sqrt{3}$:

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

since $(4\sqrt{3})^{2} = 16 \times 3 = 48$. So $7 - 4\sqrt{3}$ is a ratio­nal­is­ing fac­tor.

Exam­ple 4: a sin­gle surd that can be sim­pli­fied#

Find the sim­plest ratio­nal­is­ing fac­tor of $\sqrt{12}$.

$\sqrt{12}$ itself works, since $\sqrt{12} \times \sqrt{12} = 12$. But a smaller one exists: $\sqrt{12} \times \sqrt{3} = \sqrt{36} = 6$. So $\sqrt{3}$ is the sim­plest ratio­nal­is­ing fac­tor. The idea: mul­ti­ply by just enough to make the num­ber under the root a per­fect square.

Cube roots#

For cube roots the dif­fer­ence of squares does not help, because squar­ing a cube root does not remove it. Instead we use the sum and dif­fer­ence of cubes:

$$(x + y)(x^{2} - xy + y^{2}) = x^{3} + y^{3}$$

$$(x - y)(x^{2} + xy + y^{2}) = x^{3} - y^{3}$$

Put $x = \sqrt[3]{a}$ and $y = \sqrt[3]{b}$. Cub­ing a cube root removes it, so

$$\left(\sqrt[3]{a} + \sqrt[3]{b}\right)\left((\sqrt[3]{a})^{2} - \sqrt[3]{a} \cdot \sqrt[3]{b} + (\sqrt[3]{b})^{2}\right) = (\sqrt[3]{a})^{3} + (\sqrt[3]{b})^{3} = a + b$$

So the ratio­nal­is­ing fac­tors pair up like this:

Exam­ple 5#

Find a ratio­nal­is­ing fac­tor of $\sqrt[3]{2}$.

We need to make the num­ber under the cube root a per­fect cube. $\sqrt[3]{2} \times \sqrt[3]{4} = \sqrt[3]{8} = 2$. So $\sqrt[3]{4}$ is a ratio­nal­is­ing fac­tor.

Exam­ple 6#

Find a ratio­nal­is­ing fac­tor of $\sqrt[3]{3} + \sqrt[3]{2}$.

With $a = 3$ and $b = 2$, the ratio­nal­is­ing fac­tor is

$$(\sqrt[3]{3})^{2} - \sqrt[3]{3} \cdot \sqrt[3]{2} + (\sqrt[3]{2})^{2} = \sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}$$

and the prod­uct is $3 + 2 = 5$.

Prac­tice#

Find a ratio­nal­is­ing fac­tor of each, and the prod­uct it gives.

1. $\sqrt{5}$
2. $\sqrt{18}$ (give the sim­plest one)
3. $3 + \sqrt{2}$
4. $\sqrt{8} - \sqrt{7}$
5. $7 + 4\sqrt{3}$
6. $2\sqrt{3} - \sqrt{5}$
7. $\sqrt[3]{5}$
8. $\sqrt[3]{3} - 1$
9. Is $2\sqrt{3}$ a ratio­nal­is­ing fac­tor of $\sqrt{3}$? Explain.

Answers#