# 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:

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

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

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

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

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

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

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

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