Home Practice
For learners and parents For teachers and schools
Textbooks
Full catalogue
Pricing Support

We think you are located in United States. Is this correct?

# Test yourself now

High marks in science are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.

## End of chapter exercises

Textbook Exercise 1.8

Simplify as far as possible:

$${8}^{-\frac{2}{3}}$$

\begin{align*} 8^{-\frac{2}{3}} &= \left( 2^3 \right)^{-\frac{2}{3}} \\ &= 2^{-2} \\ &= \frac{1}{4} \end{align*}

$$\sqrt{16}+{8}^{-\frac{2}{3}}$$

\begin{align*} \sqrt{16} + 8^{-\frac{2}{3}} &= 4 + \left( 2^3 \right)^{-\frac{2}{3}} \\ &= 4 + \frac{1}{4} \\ &= 4\frac{1}{4} \end{align*}

Simplify:

$${\left({x}^{3}\right)}^{\frac{4}{3}}$$

$\left( x^3 \right)^{\frac{4}{3}} = x^4$

$${\left({s}^{2}\right)}^{\frac{1}{2}}$$

$\left( s^2 \right)^{\frac{1}{2}} = s$

$${\left({m}^{5}\right)}^{\frac{5}{3}}$$

$\left( m^5 \right)^{\frac{5}{3}} = m^{\frac{25}{3}}$

$${\left(-{m}^{2}\right)}^{\frac{4}{3}}$$

$\left(- m^2 \right)^{\frac{4}{3}} = m^{\frac{8}{3}}$

$$-{\left({m}^{2}\right)}^{\frac{4}{3}}$$

$-\left( m^2 \right)^{\frac{4}{3}} = -m^{\frac{8}{3}}$

$${\left(3{y}^{\frac{4}{3}}\right)}^{4}$$

$\left( 3y^{\frac{4}{3}} \right)^4 = 81y^{\frac{16}{3}}$

Simplify the following:

$$\dfrac{3{a}^{-2}{b}^{15}{c}^{-5}}{{\left({a}^{-4}{b}^{3}c\right)}^{\frac{-5}{2}}}$$
\begin{align*} \dfrac{3a^{-2}b^{15}c^{-5}}{(a^{-4}b^3c)^{-\frac{5}{2}}} &= \dfrac{3a^{-2}b^{15}c^{-5}}{a^{10}b^{-\frac{15}{2}}c^{-\frac{5}{2}}} \\ &= \dfrac{3b^{15+\frac{15}{2}}}{a^{12}c^{5-\frac{5}{2}}} \\ &= \dfrac{3b^{\frac{45}{2}}}{a^{12}c^{\frac{5}{2}}} \end{align*}
$${\left(9{a}^{6}{b}^{4}\right)}^{\frac{1}{2}}$$
$\left( 9a^6b^4 \right)^{\frac{1}{2}} = 3a^3b^2$
$${\left({a}^{\frac{3}{2}}{b}^{\frac{3}{4}}\right)}^{16}$$
$\left( a^{\frac{3}{2}}b^{\frac{3}{4}} \right)^{16} = a^{24}b^{12}$
$${x}^{3}\sqrt{x}$$
\begin{align*} x^3 \sqrt{x} &= x^3 \times x^{\frac{1}{2}} \\ &= x^{\frac{6}{2}} \times x^{\frac{1}{2}} \\ &= x^{\frac{7}{2}} \end{align*}
$$\sqrt[3]{{x}^{4}{b}^{5}}$$
\begin{align*} \sqrt[3]{x^4b^5} &= \left( x^4b^5 \right)^{\frac{1}{3}} \\ &= x^{\frac{4}{3}}b^{\frac{5}{3}} \end{align*}

Re-write the following expression as a power of $$x$$:

$$\dfrac{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}{x^2}$$

\begin{align*} \dfrac{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}{x^2} &= \dfrac{x \times x^{\frac{1}{2}} \times x^{\frac{1}{4}} \times x^{\frac{1}{8}} \times x^{\frac{1}{16}} }{x^2} \\ &= \dfrac{x^{\frac{31}{16}} }{x^{\frac{32}{16}}} \\ &= \dfrac{1}{x^{\frac{1}{16}}} \end{align*}

Expand:

$$\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)$$
\begin{align*} \left( \sqrt{x} - \sqrt{2} \right) \left( \sqrt{x} + \sqrt{2} \right) &= \left( \sqrt{x} \right)^2 - \left( \sqrt{2} \right)^2 \\ &= x - 2 \end{align*}

Rationalise the denominator:

$$\dfrac{10}{\sqrt{x}-1}$$
\begin{align*} \dfrac{10}{\sqrt{x} - 1} &= \dfrac{10}{\sqrt{x} - 1} \times \dfrac{\sqrt{x} + 1}{\sqrt{x} + 1} \\ &= \dfrac{10 \left( \sqrt{x} + 1 \right) }{\left( \sqrt{x} \right)^2 - 1} \\ &= \dfrac{10 \sqrt{x} + 10}{x-1} \end{align*}

Write as a single term with a rational denominator:

$$\dfrac{3}{2\sqrt{x}}+\sqrt{x}$$
\begin{align*} \dfrac{3}{2\sqrt{x}} + \sqrt{x} &= \dfrac{3 + 2\sqrt{x}\sqrt{x}}{2\sqrt{x}} \\ &= \dfrac{3 + 2x}{2\sqrt{x}} \times \frac{\sqrt{x} }{\sqrt{x} } \\ &= \dfrac{3\sqrt{x} + 2x\sqrt{x} }{2x} \end{align*}

Write in simplest surd form:

$$\sqrt{72}$$

\begin{align*} \sqrt{72} &= \sqrt{8 \times 9} \\ &= \sqrt{8} \times \sqrt{9} \\ &= 2\sqrt{2} \times 3 \\ &= 6\sqrt{2} \end{align*}

$$\sqrt{45}+\sqrt{80}$$

\begin{align*} \sqrt{45} + \sqrt{80} &= \sqrt{5 \times 9} + \sqrt{5 \times 16} \\ &= 3\sqrt{5} + 4\sqrt{5} \\ &= 7\sqrt{5} \end{align*}

$$\dfrac{\sqrt{48}}{\sqrt{12}}$$

\begin{align*} \dfrac{\sqrt{48}}{\sqrt{12}} &= \sqrt{\dfrac{48}{12}} \\ &= \sqrt{4} \\ &= 2 \\ \text{ or } \dfrac{\sqrt{48}}{\sqrt{12}} &= \dfrac{\sqrt{3 \times 16}}{\sqrt{3 \times 4}} \\ &= \dfrac{4\sqrt{3}}{2\sqrt{3}} \\ &= \frac{4}{2} \\ &= 2 \end{align*}

$$\dfrac{\sqrt{18}÷\sqrt{72}}{\sqrt{8}}$$

\begin{align*} \dfrac{\sqrt{18} \div \sqrt{72}}{\sqrt{8}} &= \dfrac{\sqrt{\frac{18}{72}}}{\sqrt{8}} \\ &= \dfrac{\sqrt{\frac{1}{4}}}{2\sqrt{2}} \\ &= \dfrac{\frac{1}{2}}{2\sqrt{2}} \\ &= \dfrac{1}{2} \times \frac{1}{2\sqrt{2}} \\ &= \dfrac{1}{4\sqrt{2}} \end{align*}

$$\dfrac{4}{\left(\sqrt{8}÷\sqrt{2}\right)}$$

\begin{align*} \dfrac{4}{\sqrt{8} \div \sqrt{2}} &= \dfrac{4}{2\sqrt{2} \div \sqrt{2}} \\ &= \dfrac{4}{2} \\ &= 2 \end{align*}

$$\dfrac{16}{\left(\sqrt{20}÷\sqrt{12}\right)}$$

\begin{align*} \dfrac{16}{\sqrt{20} \div \sqrt{12}} &= \dfrac{16}{2\sqrt{5} \div 2\sqrt{3}} \\ &= \dfrac{16}{\frac{2\sqrt{5}}{2\sqrt{3}}} \\ &= 16 \times \dfrac{\sqrt{3}}{\sqrt{5}} \\ &= 16 \times \dfrac{\sqrt{3}}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} \\ &= \dfrac{16\sqrt{15}}{5} \end{align*}

Expand and simplify:

$${\left(2+\sqrt{2}\right)}^{2}$$
\begin{align*} \left( 2 + \sqrt{2} \right)^2 &= \left( 2 + \sqrt{2} \right) \left( 2 + \sqrt{2} \right) \\ &= 4 + 4\sqrt{2} + \left( \sqrt{2} \right)^2 \\ &= 4 + 4\sqrt{2} + 2 \\ &= 6 + 4\sqrt{2} \end{align*}
$$\left(2+\sqrt{2}\right)\left(1+\sqrt{8}\right)$$
\begin{align*} \left( 2 + \sqrt{2} \right) \left( 1 + \sqrt{8} \right) &= \left( 2 + \sqrt{2} \right) \left( 1 + 2\sqrt{2} \right) \\ &= 2 + 4\sqrt{2} + \sqrt{2} + 2\left( \sqrt{2} \right)^2 \\ &= 2 + 5\sqrt{2} + 2(2) \\ &= 6 + 5\sqrt{2} \end{align*}
$$\left(1+\sqrt{3}\right)\left(1+\sqrt{8}+\sqrt{3}\right)$$
\begin{align*} \left( 1 + \sqrt{3} \right) \left( 1 + \sqrt{8} + \sqrt{3} \right) &= 1 + \sqrt{8} + \sqrt{3} + \sqrt{3} + \sqrt{8 \times 3} + \left( \sqrt{3} \right)^2\\ &= 1 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} + 3 \\ &= 4 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6} \end{align*}

Simplify, without use of a calculator:

$$\sqrt{5}\left(\sqrt{45}+2\sqrt{80}\right)$$
\begin{align*} \sqrt{5} \left( \sqrt{45} + 2\sqrt{80} \right) &= \sqrt{5} \left( \sqrt{9 \times 5} + 2\sqrt{16 \times 5} \right) \\ &= \sqrt{5} \left( 3\sqrt{5} + 8\sqrt{5} \right) \\ &= 3 \times 5 + 8 \times 5 \\ &= 15 + 40 \\ &= 55 \end{align*}
$$\dfrac{\sqrt{98}-\sqrt{8}}{\sqrt{50}}$$
\begin{align*} \dfrac{\sqrt{98} - \sqrt{8}}{\sqrt{50}} &= \dfrac{\sqrt{49 \times 2} - \sqrt{4 \times 2}}{\sqrt{25 \times 2}} \\ &= \dfrac{7\sqrt{2} - 2\sqrt{2}}{5\sqrt{2}} \\ &= \dfrac{5\sqrt{2}}{5\sqrt{2}} \\ &= 1 \end{align*}

Simplify:

$$\sqrt{98{x}^{6}}+\sqrt{128{x}^{6}}$$
\begin{align*} \sqrt{98x^6} + \sqrt{128x^6} &= \sqrt{2 \times 49 x^6} + \sqrt{2 \times 64x^6} \\ &= 7x^3 \sqrt{2} + 8x^3 \sqrt{2} \\ &= 15 \sqrt{2}x^3 \end{align*}

Rationalise the denominator:

$$\dfrac{\sqrt{5}+2}{\sqrt{5}}$$
\begin{align*} \dfrac{\sqrt{5} + 2 }{\sqrt{5}} &= \dfrac{\sqrt{5} + 2 }{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} \\ &= \dfrac{ 5 + 2\sqrt{5} }{5} \\ &= 1 + \dfrac{2\sqrt{5}}{5} \end{align*}
$$\dfrac{y-4}{\sqrt{y}-2}$$
\begin{align*} \dfrac{\sqrt{y} - 4 }{\sqrt{y} - 2 } &= \dfrac{y - 4 }{\sqrt{y} - 2 } \times \dfrac{\sqrt{y} + 2}{\sqrt{y} + 2} \\ &= \dfrac{y\sqrt{y} +2y - 4\sqrt{y} - 8}{y - 4} \\ &= \dfrac{2y + y\sqrt{y} - 4\sqrt{y} - 8}{y - 4} \end{align*}
$$\dfrac{2x-20}{\sqrt{x}-\sqrt{10}}$$
\begin{align*} \dfrac{2x - 20 }{\sqrt{x} - \sqrt{10} } &= \dfrac{2x - 20 }{\sqrt{x} - \sqrt{10} } \times \dfrac{\sqrt{x} + \sqrt{10}}{\sqrt{x} + \sqrt{10}} \\ &= \dfrac{2(x - 10) \left( \sqrt{x} + \sqrt{10} \right) }{x - 10 } \\ &= 2\sqrt{x} + 2\sqrt{10} \end{align*}

Evaluate without using a calculator: $${\left(2-\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}} \times {\left(2+\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}$$

\begin{align*} \left( 2 - \dfrac{\sqrt{7}}{2} \right)^{\frac{1}{2}} \left( 2 + \dfrac{\sqrt{7}}{2} \right)^{\frac{1}{2}} &= \left( 4 - \frac{7}{4} \right)^{\frac{1}{2}} \\ &= \sqrt{\frac{16}{4} - \frac{7}{4}}\\ &= \sqrt{\frac{9}{4}}\\ &= \frac{3}{2} \end{align*}

Prove (without the use of a calculator):

$$\sqrt{\dfrac{8}{3}}+5\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{1}{6}}=\dfrac{10\sqrt{15}+3\sqrt{6}}{6}$$
\begin{align*} \text{LHS } &= \sqrt{\dfrac{8}{3}} + 5 \sqrt{\dfrac{5}{3}} - \sqrt{\dfrac{1}{6}} \\ &= \sqrt{\frac{8}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} + 5 \sqrt{\dfrac{5}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} - \sqrt{\dfrac{1}{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\ &= \dfrac{\sqrt{8}\sqrt{3}}{3} + 5\frac{\sqrt{5}\sqrt{3}}{3} - \frac{\sqrt{6}}{6} \\ &= \dfrac{ 2\sqrt{24}}{6} + \frac{10\sqrt{15}}{6} - \frac{\sqrt{6}}{6} \\ &= \dfrac{ 4\sqrt{6} + 10\sqrt{15} - \sqrt{6}}{6} \\ &= \dfrac{ 10\sqrt{15} + 3\sqrt{6}}{6} \\ &= \text{RHS } \end{align*}

Simplify completely by showing all your steps (do not use a calculator): ${3}^{-\frac{1}{2}}\left[\sqrt{12}+\sqrt[3]{\left(3\sqrt{3}\right)}\right]$

\begin{align*} 3^{-\frac{1}{2}} \left( \sqrt{12} + \sqrt[3]{3\sqrt{3}} \right) &= \frac{1}{\sqrt{3}} \left( 2\sqrt{3} + \sqrt[3]{3\times3^{\frac{1}{2}}} \right) \\ &= \frac{1}{\sqrt{3}} \left( 2\sqrt{3} + \left( 3^{\frac{3}{2}} \right)^{\frac{1}{3}} \right) \\ &= \frac{1}{\sqrt{3}} \left( 2\sqrt{3} + 3^{\frac{1}{2}} \right) \\ &= \frac{1}{\sqrt{3}} \left( 3\sqrt{3} \right) \\ &= 3 \end{align*}

Fill in the blank surd-form number on the right hand side of the equal sign which will make the following a true statement: $$-3\sqrt{6}\times -2\sqrt{24}=-\sqrt{18}\times ...$$

\begin{align*} \text{LHS } &= -3\sqrt{6} \times -2\sqrt{24} \\ &= 6\sqrt{6}\sqrt{4 \times 6} \\ &= 12\sqrt{6}\sqrt{6} \\ &= 12(6) \\ &= 72 \\ \text{So then if LHS } &= \text{RHS } \\ \text{RHS } &= 72 \\ &= \sqrt{\text{5 184}} \\ &= -\sqrt{18} \times -\sqrt{\text{288}} \end{align*}

Solve for the unknown variable:

$$3^{x - 1} - 27 = 0$$

\begin{align*} 3^{x - 1} - 27 &= 0\\ 3^{x - 1} &= 27\\ 3^{x - 1} &= 3^3\\ x - 1 &= 3 \\ \therefore x &= 4 \end{align*}

$$8^x - \frac{1}{\sqrt[3]{8}} = 0$$

\begin{align*} 8^x - \frac{1}{\sqrt[3]{8}} &= 0\\ 2^{3x} &= \frac{1}{\sqrt[3]{2^3}} \\ 2^{3x} &= \frac{1}{2} \\ 2^{3x} &=2^{-1} \\ \therefore x &= -\frac{1}{3} \\ \text{ or } & \\ 8^x - \frac{1}{\sqrt[3]{8}} &= 0\\ 8^x &= \frac{1}{\sqrt[3]{8}} \\ 8^x &= 8^{-\frac{1}{3}}\\ \therefore x &= -\frac{1}{3} \end{align*}

$$27(4^x) = (64)3^x$$

\begin{align*} 27(4^x) &= (64)3^x \\ \frac{27}{64} &= \frac{3^x}{4^x} \\ \frac{3^3}{4^3} &= \left( \frac{3}{4} \right)^x \\ \left( \frac{3}{4} \right)^3 &= \left( \frac{3}{4} \right)^x \\ \therefore x &= 3 \end{align*}

$$\sqrt{2x - 5} = 2 - x$$

\begin{align*} \sqrt{2x - 5} &= 2 - x \\ \left( \sqrt{2x - 5} \right)^2 &= \left( 2 - x \right)^2\\ 2x - 5 &= 4 - 4x + x^2 \\ 0 &= x^2 - 6x + 9 \\ 0 &= (x - 3)(x - 3) \\ \therefore x &= 3 \\ \text{Check solution: LHS } &= \sqrt{2(3) - 5} \\ &= \sqrt{6 - 5} \\ &= \sqrt{1} \\ &= 1 \\ \text{Check solution: RHS } &= 2-3 \\ &= -1 \\ \text{RHS } &\ne \text{LHS }\\ \therefore &\text{No solution} \end{align*}

$$2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 = 0$$

\begin{align*} 2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 &= 0 \\ \left( 2x^{\frac{1}{3}} - 1 \right) \left( x^{\frac{1}{3}} + 2 \right) &= 0 \\ \therefore 2x^{\frac{1}{3}} - 1 &= 0 \\ 2x^{\frac{1}{3}} &= 1 \\ x^{\frac{1}{3}} &= \frac{1}{2} \\ \left( x^{\frac{1}{3}} \right)^3 &= \left( \frac{1}{2} \right)^3 \\ \therefore x &= \frac{1}{8} \\ \text{ or } & \\ x^{\frac{1}{3}} + 2 &= 0 \\ x^{\frac{1}{3}} &= -2 \\ \left( x^{\frac{1}{3}} \right)^3 &= \left( -2 \right)^3 \\ x &= -8 \\ \text{Therefore } x = \frac{1}{8} &\text{ or } x = -8 \end{align*}

Show that $$\sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3}$$ is equal to $$\text{3}$$

\begin{align*} \text{LHS } &= \sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} \\ &= \sqrt{\dfrac{3^{x+1}}{3^{x-1}} - \dfrac{3^{x}}{3^{x-1}}+ 3} \\ &= \sqrt{3^{x+1-x+1} - 3^{x-x+1} + 3} \\ &= \sqrt{3^{2} - 3^{1} + 3} \\ &= \sqrt{3^{2}} \\ &= 3 \\ &= \text{RHS } \end{align*}

Hence solve $$\sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} = \left( \dfrac{1}{3} \right)^{x-2}$$

\begin{align*} \sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} &= \left( \dfrac{1}{3} \right)^{x-2} \\ 3 &= \left( \frac{1}{3} \right)^{x-2} \\ 3 &= \left( 3\right)^{-x+2} \\ 1 &= -x + 2\\ x &= 1 \end{align*}