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7. Rationalising homework (28 Sep)

The rationalising exercises for Monday 28 September, with the solution to each part folded away.

Homework for Monday 28 September

Each part has its solution folded away. Do the exercise in your notebook and, once you have it, click on the statement to unfold the solution and check it.

The three types of rationalising are in section 6.

Rationalising

1 Rationalise the following fractions
a) $\dfrac{1}{\sqrt{3}}$

Type 1: multiply and divide by $\sqrt{3}$.

$$\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$
b) $\dfrac{3}{\sqrt[4]{2^{3}}}$

Type 2: $3 + p = 4$, so $p = 1$.

$$\frac{3}{\sqrt[4]{2^{3}}} \cdot \frac{\sqrt[4]{2}}{\sqrt[4]{2}} = \frac{3\sqrt[4]{2}}{\sqrt[4]{2^{4}}} = \frac{3\sqrt[4]{2}}{2}$$
c) $\dfrac{2}{\sqrt[3]{5^{2}}}$

Type 2: $2 + p = 3$, so $p = 1$.

$$\frac{2}{\sqrt[3]{5^{2}}} \cdot \frac{\sqrt[3]{5}}{\sqrt[3]{5}} = \frac{2\sqrt[3]{5}}{\sqrt[3]{5^{3}}} = \frac{2\sqrt[3]{5}}{5}$$
d) $\dfrac{5}{\sqrt[7]{3^{5}}}$

Type 2: $5 + p = 7$, so $p = 2$.

$$\frac{5}{\sqrt[7]{3^{5}}} \cdot \frac{\sqrt[7]{3^{2}}}{\sqrt[7]{3^{2}}} = \frac{5\sqrt[7]{3^{2}}}{\sqrt[7]{3^{7}}} = \frac{5\sqrt[7]{9}}{3}$$
2 Rationalise the following expressions
a) $\dfrac{2}{\sqrt{3} - \sqrt{5}}$

Type 3: the conjugate of $\sqrt{3} - \sqrt{5}$ is $\sqrt{3} + \sqrt{5}$.

$$\frac{2}{\sqrt{3} - \sqrt{5}} \cdot \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \frac{2(\sqrt{3} + \sqrt{5})}{(\sqrt{3})^{2} - (\sqrt{5})^{2}} = \frac{2(\sqrt{3} + \sqrt{5})}{3 - 5}$$
$$= \frac{2(\sqrt{3} + \sqrt{5})}{-2} = -\sqrt{3} - \sqrt{5}$$
b) $\dfrac{5}{3\sqrt{3} - \sqrt{2}}$

Type 3: the conjugate of $3\sqrt{3} - \sqrt{2}$ is $3\sqrt{3} + \sqrt{2}$.

$$\frac{5}{3\sqrt{3} - \sqrt{2}} \cdot \frac{3\sqrt{3} + \sqrt{2}}{3\sqrt{3} + \sqrt{2}} = \frac{5(3\sqrt{3} + \sqrt{2})}{(3\sqrt{3})^{2} - (\sqrt{2})^{2}} = \frac{5(3\sqrt{3} + \sqrt{2})}{27 - 2}$$
$$= \frac{5(3\sqrt{3} + \sqrt{2})}{25} = \frac{3\sqrt{3} + \sqrt{2}}{5}$$

Careful: $(3\sqrt{3})^{2} = 3^{2} \cdot 3 = 27$. The $3$ outside is squared too.

c) $\dfrac{-3}{5 - 4\sqrt{2}}$

Type 3: the conjugate of $5 - 4\sqrt{2}$ is $5 + 4\sqrt{2}$.

$$\frac{-3}{5 - 4\sqrt{2}} \cdot \frac{5 + 4\sqrt{2}}{5 + 4\sqrt{2}} = \frac{-3(5 + 4\sqrt{2})}{5^{2} - (4\sqrt{2})^{2}} = \frac{-15 - 12\sqrt{2}}{25 - 32}$$
$$= \frac{-15 - 12\sqrt{2}}{-7} = \frac{15 + 12\sqrt{2}}{7}$$

$(4\sqrt{2})^{2} = 16 \cdot 2 = 32$, and the denominator comes out negative: the two minus signs cancel.

d) $\dfrac{2}{\sqrt{3} + \sqrt{5} + 3}$

With three terms, group the two roots, $(\sqrt{3} + \sqrt{5}) + 3$, and do it in two steps. First, by the conjugate $(\sqrt{3} + \sqrt{5}) - 3$:

$$\frac{2}{(\sqrt{3} + \sqrt{5}) + 3} \cdot \frac{(\sqrt{3} + \sqrt{5}) - 3}{(\sqrt{3} + \sqrt{5}) - 3} = \frac{2(\sqrt{3} + \sqrt{5} - 3)}{(\sqrt{3} + \sqrt{5})^{2} - 3^{2}}$$
$$= \frac{2(\sqrt{3} + \sqrt{5} - 3)}{2\sqrt{15} - 1}$$

In the denominator, $(\sqrt{3} + \sqrt{5})^{2} = 3 + 2\sqrt{15} + 5$. There is still a root, so now by the conjugate of $2\sqrt{15} - 1$, which is $2\sqrt{15} + 1$:

$$\frac{2(\sqrt{3} + \sqrt{5} - 3)}{2\sqrt{15} - 1} \cdot \frac{2\sqrt{15} + 1}{2\sqrt{15} + 1} = \frac{2(\sqrt{3} + \sqrt{5} - 3)(2\sqrt{15} + 1)}{(2\sqrt{15})^{2} - 1^{2}}$$
$$= \frac{2(\sqrt{3} + \sqrt{5} - 3)(2\sqrt{15} + 1)}{59}$$

Multiply out the numerator term by term, using $\sqrt{45} = 3\sqrt{5}$ and $\sqrt{75} = 5\sqrt{3}$:

$$(\sqrt{3} + \sqrt{5} - 3)(2\sqrt{15} + 1) = 2\sqrt{45} + \sqrt{3} + 2\sqrt{75} + \sqrt{5} - 6\sqrt{15} - 3$$
$$= 11\sqrt{3} + 7\sqrt{5} - 6\sqrt{15} - 3$$

Which leaves:

$$\frac{2}{\sqrt{3} + \sqrt{5} + 3} = \frac{22\sqrt{3} + 14\sqrt{5} - 12\sqrt{15} - 6}{59}$$

Calculator check: the statement and the result both give $0.287\ldots$