Roots and Powers: Quick Revision

A brief note on Roots and Powers. Read it before a test. Try each question at the end before you open its answer.

Rules for square roots

For positive $a$ and $b$:

• $\sqrt{a}\,\sqrt{b} = \sqrt{ab}$ and $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$.
• $\sqrt{a} + \sqrt{b}$ is not $\sqrt{a + b}$.
• Simplify by taking out a perfect square: $\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$.
• $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$, and $(\sqrt{a} \pm \sqrt{b})^2 = a + b \pm 2\sqrt{ab}$.

Rationalising

Laws of exponents

$a^m a^n = a^{m+n}$, $(a^m)^n = a^{mn}$, $\dfrac{a^m}{a^n} = a^{m-n}$, $a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$.

$a^{1/n}$ is the $n$-th root of $a$, and $a^{m/n} = \left(\sqrt[n]{a}\right)^m$. Take the root first.

A worked example

Rationalise $\dfrac{4}{\sqrt{6} - \sqrt{2}}$.

1. The conjugate. Multiply the top and the bottom by $\sqrt{6} + \sqrt{2}$.
2. The bottom. $6 - 2 = 4$.
3. Simplify. $\dfrac{4(\sqrt{6} + \sqrt{2})}{4} = \sqrt{6} + \sqrt{2}$.

Practice with new numbers

A program to try

The program checks the worked example with decimals.

Mistakes to avoid

• Adding roots as if they were numbers. $\sqrt{9} + \sqrt{16} = 7$, not $5$.
• Forgetting the middle term. $(\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}$.
• Multiplying only the bottom. The top must be multiplied by the same number.
• Reading $a^{-n}$ as a negative number. $2^{-2} = \dfrac{1}{4}$.

Check yourself

1. Simplify $\sqrt{45} + \sqrt{20}$.

2. Rationalise $\dfrac{3}{\sqrt{6}}$.

3. Find $1000^{2/3}$.

4. Find $9^{-3/2}$.