Indices and surds

Unit 1 Indices and surds

Key unit competence: Use indices and surds to simplify mathematical expressions from real-life situations.

Key ideas

• In $a^n$, $a$ is the base and $n$ the index (power, exponent): $a^n=a\times a\times \cdots \times a$ ($n$ factors).
• Laws of indices: $a^m\times a^n=a^{m+n}$, $a^m÷a^n=a^{m-n}$, $(a^m{)}^n=a^{mn}$, $(ab{)}^n=a^n{b}^n$, $a^0=1$ ($a\ne 0$), $a^{-n}=\frac{1}{a^n}$.
• Fractional indices: $a^{\frac{1}{n}}=\sqrt[n]{a}$ and $a^{\frac{m}{n}}={\left(\sqrt[n]{a}\right)}^m$.
• Standard form (scientific notation): $A\times 10^n$ with $1\le A<10$ and $n$ an integer.
• A surd is a root that cannot be simplified to a rational number, e.g. $\sqrt{2}$, $\sqrt{5}$. Use $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ to simplify: $\sqrt{12}=\sqrt{4\times 3}=2\sqrt{3}$.
• Only like surds can be added: $3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$, but $\sqrt{2}+\sqrt{3}$ cannot be simplified.
• Rationalising the denominator: $\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{b}$; for $\frac{1}{a+\sqrt{b}}$ multiply top and bottom by the conjugate $a-\sqrt{b}$, using $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$.

Formulas to know

Common misconceptions

• "$2^3\times 2^4=4^7$." No: keep the base and add the powers: $2^7$.
• "$\sqrt{9+16}=\sqrt{9}+\sqrt{16}=7$." No: $\sqrt{25}=5$. The root of a sum is not the sum of the roots.
• "$a^0=0$." No: any non-zero number to the power 0 equals 1.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Simplify $\frac{x^5{y}^3}{x^{2}y}$.

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   $x^3{y}^2$.

   Common mistake: Dividing the indices instead of subtracting them.

2. Estimate $\sqrt{30}$ to one decimal place, then check by squaring.

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   $5^2=25$ and $6^2=36$, so it is about 5.5: $5.5^2=30.25$, $5.4^2=29.16$. Hence $\sqrt{30}\approx 5.5$.

   Common mistake: Taking the halfway value without checking.

3. Evaluate $16^{\frac{3}{4}}$ and $16^{-\frac{1}{2}}$.

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   $16^{\frac{3}{4}}=(\sqrt[4]{16}{)}^3=2^3=8$; $16^{-\frac{1}{2}}=\frac{1}{4}$.

   Common mistake: Multiplying 16 by $\frac{3}{4}$.

4. Solve $5^{2x-1}=125$.

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   $125=5^3$, $2x-1=3$, $x=2$.

   Common mistake: Equating $2x-1$ with 125.

5. The mass of the Earth is about 5 970 000 000 000 000 000 000 000 kg. Write it in standard form.

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   $5.97\times 10^{24}$ kg.

   Common mistake: Miscounting the number of places.

6. Rationalise $\frac{1}{\sqrt{5}-\sqrt{2}}$.

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   $\frac{\sqrt{5}+\sqrt{2}}{5-2}=\frac{\sqrt{5}+\sqrt{2}}{3}$.

   Common mistake: Using $\sqrt{5}-\sqrt{2}$ again instead of the conjugate.

7. Simplify $2\sqrt{27}+\sqrt{12}$.

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   $6\sqrt{3}+2\sqrt{3}=8\sqrt{3}$.

   Common mistake: Adding the numbers under the roots.

8. Rationalise $\frac{6}{\sqrt{3}}$.

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   $\frac{6\sqrt{3}}{3}=2\sqrt{3}$.

   Common mistake: Multiplying only the denominator by $\sqrt{3}$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $\sqrt{75}-2\sqrt{12}+\sqrt{27}$.	a. $x=4$
   2. Simplify $3\sqrt{2}+5\sqrt{2}$.	b. $8\sqrt{2}$
   3. Solve $2^{x+1}=32$.	c. $4\sqrt{3}$
   4. Rationalise $\frac{6}{3-\sqrt{3}}$.	d. $3+\sqrt{3}$

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   1 → c ($4\sqrt{3}$) · 2 → b ($8\sqrt{2}$) · 3 → a ($x=4$) · 4 → d ($3+\sqrt{3}$)

   • Step 1. $\sqrt{75}=5\sqrt{3}$, $2\sqrt{12}=4\sqrt{3}$, $\sqrt{27}=3\sqrt{3}$.
     Step 2. $5\sqrt{3}-4\sqrt{3}+3\sqrt{3}=4\sqrt{3}$.
   • Like surds: add the coefficients.
   • Step 1. $32=2^5$, so $2^{x+1}=2^5$.
     Step 2. $x+1=5$, so $x=4$.
   • Step 1. Multiply by the conjugate: $\frac{6(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{6(3+\sqrt{3})}{9-3}$.
     Step 2. $\frac{6(3+\sqrt{3})}{6}=3+\sqrt{3}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $x=\frac{3}{2}$  ·  $4.5\times 10^4$  ·  4  ·  2
   1. Evaluate $8^{\frac{2}{3}}$. →
   2. Solve $9^x=27$. →
   3. Write 45 000 in standard form. →
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   1: 4 · 2: $x=\frac{3}{2}$ · 3: $4.5\times 10^4$

   • Step 1. $\sqrt[3]{8}=2$.
     Step 2. Then $2^2=4$.
   • Step 1. $3^{2x}=3^3$.
     Step 2. So $2x=3$.
   • $1\le 4.5<10$ and the decimal point moves 4 places.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Simplify $\sqrt{50}$.	a. $3x^{2}y$
   2. Evaluate ${\left(\frac{16}{81}\right)}^{-\frac{3}{4}}$.	b. $5\sqrt{2}$
   3. Simplify $(2x^3{)}^2$.	c. $\frac{27}{8}$
   4. Simplify $\frac{(3x^{2}y{)}^3}{9x^4{y}^2}$.	d. $4x^6$

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   1 → b ($5\sqrt{2}$) · 2 → c ($\frac{27}{8}$) · 3 → d ($4x^6$) · 4 → a ($3x^{2}y$)

   • $\sqrt{25\times 2}=5\sqrt{2}$.
   • Step 1. A negative power flips the fraction: ${\left(\frac{81}{16}\right)}^{\frac{3}{4}}$.
     Step 2. Fourth root: $\frac{3}{2}$; cube it: $\frac{27}{8}$.
   • $(2{)}^2(x^3{)}^2=4x^6$.
   • Step 1. $(3x^{2}y{)}^3=27x^6{y}^3$.
     Step 2. $\frac{27x^6{y}^3}{9x^4{y}^2}=3x^{6-4}{y}^{3-2}=3x^{2}y$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\sqrt{2}$  ·  5  ·  $\frac{1}{3}$  ·  $5\sqrt{2}$
   1. Evaluate $25^{\frac{1}{2}}$. →
   2. Evaluate $27^{-\frac{1}{3}}$. →
   3. Simplify $\sqrt{18}-\sqrt{8}$. →
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   1: 5 · 2: $\frac{1}{3}$ · 3: $\sqrt{2}$

   • Power $\frac{1}{2}$ means square root.
   • $27^{\frac{1}{3}}=3$, and the negative index gives the reciprocal.
   • $3\sqrt{2}-2\sqrt{2}=\sqrt{2}$.

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