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

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

Solution:

Question 10:

Solution:

Question 11:

Solution:

Question 12:

Solution:

Question 13:

Solution:

Question 14:

Solution:

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

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.

Solution:

$\begin{array}{l}\sqrt[3]{4\frac{17}{27}}÷\sqrt{1\frac{9}{16}}=\sqrt[3]{\frac{\overline{)125}}{27}}÷\sqrt{\frac{25}{16}}\\ \text{}=\frac{5}{3}÷\frac{5}{4}\\ \text{}=\frac{\overline{)5}}{3}×\frac{4}{\overline{)5}}\\ \text{}=\overline{)\frac{4}{3}}\end{array}$

Question 7:
Complete the operation steps below by filling in the boxes using suitable numbers.
$\begin{array}{l}{\left(\sqrt{2\frac{7}{9}}-\sqrt[3]{\frac{27}{64}}\right)}^{2}={\left(\sqrt{\frac{\overline{)}}{9}}-\frac{3}{\overline{)}}\right)}^{2}\\ \text{}={\left(\frac{\overline{)}}{12}\right)}^{2}\\ \text{}=\overline{)}\end{array}$

Solution:
$\begin{array}{l}{\left(\sqrt{2\frac{7}{9}}-\sqrt[3]{\frac{27}{64}}\right)}^{2}={\left(\sqrt{\frac{\overline{)25}}{9}}-\frac{3}{\overline{)4}}\right)}^{2}\\ \text{}={\left(\frac{5}{3}-\frac{3}{4}\right)}^{2}\\ \text{}={\left(\frac{\left(5×4\right)-\left(3×3\right)}{12}\right)}^{2}\\ \text{}={\left(\frac{\overline{)11}}{12}\right)}^{2}\\ \text{}=\overline{)\frac{121}{144}}\end{array}$

Question 8:
Complete the operation steps below by filling in the boxes using suitable numbers.
$\begin{array}{l}\sqrt[3]{1\frac{61}{64}}-{0.3}^{2}=\sqrt[3]{\frac{\overline{)}}{64}}-{0.3}^{2}\\ \text{}=\frac{\overline{)}}{4}-\overline{)}\\ \text{}=\overline{)}\end{array}$

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

Question 9:
Find the value of:
$\begin{array}{l}\text{(a)}\sqrt[3]{\frac{10}{27}-5}\\ \text{(b)}\sqrt{\frac{4}{49}}×\sqrt[3]{-0.216}\end{array}$

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

$\begin{array}{l}\text{(b)}\sqrt{\frac{4}{49}}×\sqrt[3]{-0.216}=\frac{2}{7}×\sqrt[3]{\frac{216}{1000}}\\ \text{}=\frac{\overline{)2}}{7}×\frac{6}{5\overline{)10}}\\ \text{}=\frac{6}{35}\end{array}$

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

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

Question 1:
Calculate the values of the following:
$\begin{array}{l}\text{(a)}\sqrt{\frac{50}{98}}\\ \text{(b)}\sqrt{1\frac{17}{64}}\\ \text{(c)}\sqrt{81}-\sqrt{0.01}\\ \text{(d)}\sqrt{3.24}\end{array}$

Solution:
$\text{(a)}\sqrt{\frac{50}{98}}=\sqrt{\frac{\overline{)50}25}{\overline{)98}49}}=\sqrt{\frac{25}{49}}=\frac{5}{7}$

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

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

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

Question 2:
Calculate the values of the following:
$\begin{array}{l}\text{(a)}\sqrt[3]{\frac{16}{250}}\\ \text{(b)}\sqrt[3]{-\frac{4}{256}}\\ \text{(c)}\sqrt[3]{0.008}\\ \text{(d)}\sqrt[3]{0.729}\end{array}$

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

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

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

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

Question 3:
Find the value of $\sqrt[3]{3\frac{3}{8}}+\sqrt{2\frac{14}{25}}.$

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

Question 4:
Find the values of the following:
(a) 1 – (–0.3)3.
(b) ${\left(2.1÷\sqrt[3]{27}\right)}^{2}$

Solution:
(a)
1 – (–0.3)3 = 1 – [(–0.3) × (–0.3) × (–0.3)]
= 1 – (–0.027)
= 1 + 0.027
= 1.027

(b)
$\begin{array}{l}{\left(2.1÷\sqrt[3]{27}\right)}^{2}={\left(2.1÷3\right)}^{2}\\ \text{}={\left(0.7\right)}^{2}\\ \text{}=0.49\end{array}$

Question 5:
Find the values of the following:
$\begin{array}{l}\text{(a)}{\left(9+\sqrt[3]{-8}\right)}^{2}\\ \text{(b)}\sqrt{144}÷\sqrt[3]{216}×{0.3}^{3}\end{array}$

Solution:
$\begin{array}{l}\text{(a)}{\left(9+\sqrt[3]{-8}\right)}^{2}={\left[9+\left(-2\right)\right]}^{2}\\ \text{}={7}^{2}\\ \text{}=49\end{array}$

$\begin{array}{l}\text{(b)}\sqrt{144}÷\sqrt[3]{216}×{0.3}^{3}\\ \text{}=144÷6×\left(0.3×0.3×0.3\right)\\ \text{}=24×0.027\\ \text{}=0.648\end{array}$

2.1 Squares, Square Roots, Cube and Cube Roots

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) 13= 13 × 13 = 169
(b)   (–10)= (–10) × (–10) = 100
(c) (0.4)2 = 0.4 × 0.4 = 0.16
(d)   (–0.06)= (–0.06) × (–0.06) = 0.0036
$\begin{array}{l}\text{(e)}{\left(3\frac{1}{2}\right)}^{2}={\left(\frac{7}{2}\right)}^{2}=\frac{7}{2}×\frac{7}{2}=\frac{49}{4}\\ \left(\text{f}\right)\text{}{\left(-1\frac{2}{7}\right)}^{2}={\left(-\frac{9}{7}\right)}^{2}=\left(-\frac{9}{7}\right)×\left(-\frac{9}{7}\right)=\frac{81}{49}\end{array}$

(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:
$\begin{array}{l}\text{(a)}\sqrt{169}=\sqrt{13×13}=13\\ \text{(b)}\sqrt{\frac{25}{64}}=\sqrt{\frac{5×5}{8×8}}=\frac{5}{8}\\ \text{(c)}\sqrt{\frac{72}{98}}=\sqrt{\frac{\overline{)72}36}{\overline{)98}49}}=\sqrt{\frac{6×6}{7×7}}=\frac{6}{7}\\ \text{(d)}\sqrt{3\frac{1}{16}}=\sqrt{\frac{49}{16}}=\frac{7}{4}=1\frac{3}{4}\\ \text{(e)}\sqrt{1.44}=\sqrt{1\frac{\overline{)44}11}{\overline{)100}25}}=\sqrt{\frac{36}{25}}=\frac{6}{5}=1\frac{1}{5}\end{array}$

(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:
$\begin{array}{l}\sqrt[3]{64}=\sqrt[3]{4×4×4}\\ \text{}=4\end{array}$
$\sqrt[3]{64}$ is read as ‘cube root of sixty-four’.

2.
The cube root of a positive number is positive.
Example:
$\begin{array}{l}\sqrt[3]{125}=\sqrt[3]{5×5×5}\\ \text{}=5\end{array}$

3.
The cube root of a negative number is negative.
Example:
$\begin{array}{l}\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)×\left(-5\right)×\left(-5\right)}\\ \text{}=-5\end{array}$

4.
To determine the cube roots of fractions, the fractions should be simplified to numerators and denominators that are cubes of integers.
Example:
$\begin{array}{l}\sqrt[3]{\frac{16}{250}}=\sqrt[3]{\frac{\overline{)16}8}{\overline{)250}125}}\\ \text{}=\sqrt[3]{\frac{8}{125}}\\ \text{}=\frac{2}{5}\end{array}$