2.2.4 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Posted on June 17, 2017 by user

Question 15:
Match each of the following with the correct value.

Solution:

2.2.3 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Posted on June 16, 2017 by user

Question 10:

\begin{array}{l} (a) Find the value of \sqrt{0.09} \\ (b) Calculate the value of {(\frac{5}{7} \times \sqrt{49})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3 \\ (b) \\ {(\frac{5}{7} \times \sqrt{49})}^{2} = {(\frac{5}{7} \times 7)}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 11:

\begin{array}{l} (a) Find the value of \sqrt[3]{- \frac{1}{27}} \\ (b) Calculate the value of {(7 + \sqrt[3]{- 8})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{- \frac{1}{27}} = \sqrt[3]{- \frac{1}{3} \times - \frac{1}{3} \times - \frac{1}{3}} = - \frac{1}{3} \\ (b) \\ {(7 + \sqrt[3]{- 8})}^{2} = {[7 + (- 2)]}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 12:

\begin{array}{l} (a) Find the value of {(- \frac{1}{4})}^{3} . \\ (b) Calculate the value of {(4.2 \div \sqrt[3]{27})}^{2} . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{1}{4})}^{3} = - \frac{1}{4} \times - \frac{1}{4} \times - \frac{1}{4} \\ = - \frac{1}{64} \\ (b) \\ {(4.2 \div \sqrt[3]{27})}^{2} = {(4.2 \div 3)}^{2} \\ = {1.4}^{2} \\ = 1.4 \times 1.4 \\ = 1.96 \end{array}

Question 13:

\begin{array}{l} (a) Find the value of \sqrt[3]{0.008} . \\ (b) Calculate the value of 3^{2} \times \sqrt[3]{- \frac{64}{27}} . \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \\ (b) \\ 3^{2} \times \sqrt[3]{- \frac{64}{27}} = 9^{3} \times - \frac{4}{3} \\ = 12 \end{array}

Question 14:

\begin{array}{l} (a) Find the value of {(- \frac{3}{4})}^{3} . \\ (b) Calculate the value of \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{3}{4})}^{3} = - \frac{3}{4} \times - \frac{3}{4} \times - \frac{3}{4} \\ = - \frac{27}{64} \\ (b) \\ \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) = \sqrt[3]{\frac{64}{27}} \div (4 - 9) \\ = \frac{4}{3} \div - 5 \\ = \frac{4}{3} \times - \frac{1}{5} \\ = - \frac{4}{15} \end{array}

2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Posted on June 14, 2017 by user

2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Question 6:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{}{3} \div \frac{5}{4} \\ = \frac{}{3} \times \frac{4}{5} \\ = \end{array}

Solution:

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{125}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{5}{3} \div \frac{5}{4} \\ = \frac{5}{3} \times \frac{4}{5} \\ = \frac{4}{3} \end{array}

Question 7:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{}{9}} - \frac{3}{})}^{2} \\ = {(\frac{}{12})}^{2} \\ = \end{array}

Solution:

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{25}{9}} - \frac{3}{4})}^{2} \\ = {(\frac{5}{3} - \frac{3}{4})}^{2} \\ = {(\frac{(5 \times 4) - (3 \times 3)}{12})}^{2} \\ = {(\frac{11}{12})}^{2} \\ = \frac{121}{144} \end{array}

Question 8:

Complete the operation steps below by filling in the boxes using suitable numbers.

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{}{64}} - {0.3}^{2} \\ = \frac{}{4} - \\ = \end{array}

Solution:

\begin{array}{l} \sqrt[3]{1 \frac{61}{64}} - {0.3}^{2} = \sqrt[3]{\frac{125}{64}} - {0.3}^{2} \\ = \frac{5}{4} - 0.09 \\ = 1.25 - 0.09 \\ = 1.16 \end{array}

Question 9:

Find the value of:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} \\ (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} \end{array}

Solution:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} = \sqrt[3]{\frac{10 - 135}{27}} \\ = \sqrt[3]{- \frac{125}{27}} \\ = - \frac{5}{3} \end{array}

\begin{array}{l} (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} = \frac{2}{7} \times \sqrt[3]{\frac{216}{1000}} \\ = \frac{2}{7} \times \frac{6}{510} \\ = \frac{6}{35} \end{array}

2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

Posted on June 12, 2017 by user

2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

Question 1:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt{\frac{50}{98}} \\ (b) \sqrt{1 \frac{17}{64}} \\ (c) \sqrt{81} - \sqrt{0.01} \\ (d) \sqrt{3.24} \end{array}

Solution:

(a) \sqrt{\frac{50}{98}} = \sqrt{\frac{5025}{9849}} = \sqrt{\frac{25}{49}} = \frac{5}{7}

(b) \sqrt{1 \frac{17}{64}} = \sqrt{\frac{81}{64}} = \frac{9}{8} = 1 \frac{1}{8}

\begin{array}{l} (c) \sqrt{81} - \sqrt{0.01} = 9 - \sqrt{\frac{1}{100}} \\ = 9 - \frac{1}{10} \\ = 9 - 0.1 \\ = 8.9 \end{array}

\begin{array}{l} (d) \sqrt{3.24} = \sqrt{3 \frac{24}{100}} = \sqrt{3 \frac{6}{25}} \\ = \sqrt{\frac{81}{25}} \\ = \frac{9}{5} = 1 \frac{4}{5} \end{array}

Question 2:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt[3]{\frac{16}{250}} \\ (b) \sqrt[3]{- \frac{4}{256}} \\ (c) \sqrt[3]{0.008} \\ (d) \sqrt[3]{0.729} \end{array}

Solution:

(a) \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{8}{125}} = \frac{2}{5}

(b) \sqrt[3]{- \frac{4}{256}} = \sqrt[3]{- \frac{1}{64}} = - \frac{1}{4}

\begin{array}{l} (c) \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \end{array}

\begin{array}{l} (d) \sqrt[3]{- 0.729} = \sqrt[3]{- \frac{729}{1000}} \\ = - \frac{9}{10} \\ = - 0.9 \end{array}

Question 3:

Find the value of

\sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} .

Solution:

\begin{array}{l} \sqrt[3]{3 \frac{3}{8}} + \sqrt{2 \frac{14}{25}} = \sqrt[3]{\frac{27}{8}} + \sqrt{\frac{64}{25}} \\ = \frac{3}{2} + \frac{8}{5} \\ = \frac{31}{10} = 3 \frac{1}{10} \end{array}

Question 4:

Find the values of the following:

(a) 1 – (–0.3)³.

(b)

{(2.1 \div \sqrt[3]{27})}^{2}

Solution:

(a)

1 – (–0.3)³= 1 – [(–0.3) × (–0.3) × (–0.3)]

= 1 – (–0.027)

= 1 + 0.027

= 1.027

(b)

\begin{array}{l} {(2.1 \div \sqrt[3]{27})}^{2} = {(2.1 \div 3)}^{2} \\ = {(0.7)}^{2} \\ = 0.49 \end{array}

Question 5:

Find the values of the following:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} \\ (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \end{array}

Solution:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} = {[9 + (- 2)]}^{2} \\ = 7^{2} \\ = 49 \end{array}

\begin{array}{l} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \\ = 144 \div 6 \times (0.3 \times 0.3 \times 0.3) \\ = 24 \times 0.027 \\ = 0.648 \end{array}

2.1 Squares, Square Roots, Cube and Cube Roots

Posted on June 10, 2017 by user

2.1 Squares, Square Roots, Cube and Cube Roots

(A) Squares

The square of a number is the answer you get when you multiply a number by itself.

Example:

(a) 13²= 13 × 13 = 169

(b) (–10)²= (–10) × (–10) = 100

(d) (–0.06)²= (–0.06) × (–0.06) = 0.0036

\begin{array}{l} (e) {(3 \frac{1}{2})}^{2} = {(\frac{7}{2})}^{2} = \frac{7}{2} \times \frac{7}{2} = \frac{49}{4} \\ (f) {(- 1 \frac{2}{7})}^{2} = {(- \frac{9}{7})}^{2} = (- \frac{9}{7}) \times (- \frac{9}{7}) = \frac{81}{49} \end{array}

(B) Perfect Squares

1. Perfect squares are the squares of whole numbers.

2. Perfect squares are formed by multiplying a whole number by itself.

Example:

4 = 2 × 2 9 = 3 × 3 16 = 4 × 4

3. The first twelve perfect squares are:

= 1², 2², 3², 4², 5², 6², 7², 8², 9², 10², 11², 12²

= 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

(C) Square Roots

1. The square root of a positive number is a number multiplied by itself whose product is equal to the given number.

Example:

\begin{array}{l} (a) \sqrt{169} = \sqrt{13 \times 13} = 13 \\ (b) \sqrt{\frac{25}{64}} = \sqrt{\frac{5 \times 5}{8 \times 8}} = \frac{5}{8} \\ (c) \sqrt{\frac{72}{98}} = \sqrt{\frac{7236}{9849}} = \sqrt{\frac{6 \times 6}{7 \times 7}} = \frac{6}{7} \\ (d) \sqrt{3 \frac{1}{16}} = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1 \frac{3}{4} \\ (e) \sqrt{1.44} = \sqrt{1 \frac{4411}{10025}} = \sqrt{\frac{36}{25}} = \frac{6}{5} = 1 \frac{1}{5} \end{array}

(D) Cubes

1. The cube of a number is obtained when that number is multiplied by itself twice.

Example:

The cube of 3 is written as

3³ = 3 × 3 × 3

= 27

2. The cube of a negative number is negative.

Example:

(–2)³ = (–2) × (–2) × (–2)

= –8

3. The cube of zero is zero. The cube of one is one, 1³ = 1.

(E) Cube Roots

1. The cube root of a number is a number which, when multiplied by itself twice, produces the particular number.

" \sqrt[3]{} "

is the symbol for cube root.

Example:

\begin{array}{l} \sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4} \\ = 4 \end{array}

\sqrt[3]{64}

is read as ‘cube root of sixty-four’.

2. The cube root of a positive number is positive.

Example:

\begin{array}{l} \sqrt[3]{125} = \sqrt[3]{5 \times 5 \times 5} \\ = 5 \end{array}

3. The cube root of a negative number is negative.

Example:

\begin{array}{l} \sqrt[3]{- 125} = \sqrt[3]{(- 5) \times (- 5) \times (- 5)} \\ = - 5 \end{array}

4. To determine the cube roots of fractions, the fractions should be simplified to numerators and denominators that are cubes of integers.

Example:

\begin{array}{l} \sqrt[3]{\frac{16}{250}} = \sqrt[3]{\frac{168}{250125}} \\ = \sqrt[3]{\frac{8}{125}} \\ = \frac{2}{5} \end{array}