Exponents and Radicals : Exercise (2.4) - Solution

1. Simplify the following.
   
   (a) $3 \sqrt{5}+7 \sqrt{5}$
   
   
   
   

   $ \begin{aligned} &3\sqrt{5}+7\sqrt{5}\\\\ =&10\sqrt{5}\end{aligned}$

   
   
   (b) $\sqrt{75}-\sqrt{12}$
   
   
   
   

   $\begin{aligned} &\sqrt{{75}}-\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-2\sqrt{3}\\\\ =&3\sqrt{3} \end{aligned}$

   
   
   (c) $3 \cdot 3 \sqrt{3} \cdot 3 \sqrt{27}$
   
   
   
   

   $\begin{aligned} &3\cdot 3\sqrt{3}\cdot 3\sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{27}}\\\\ =&27\cdot \sqrt{3}\cdot \sqrt{{9\times 3}}\\\\ =&27\cdot \sqrt{3}\cdot 3\sqrt{3}\\\\ =&81\cdot \sqrt{3}\cdot \sqrt{3}\\\\ =&81\cdot 3\\\\ =&243 \end{aligned}$

   
   
   (d) $2 \sqrt{5} \cdot 3 \sqrt{2}$
   
   
   
   

   $\begin{aligned} &2 \sqrt{5} \cdot 3 \sqrt{2}\\\\ =&6\sqrt{10} \end{aligned}$

   
   
   (e) $(4-\sqrt{3})^{2}$
   
   
   
   

   $\begin{aligned} &(4-\sqrt{3})^2\\\\ =&16-2\times 4\times \sqrt{3}+3\\\\ =&19-8\sqrt{3} \end{aligned}$

   
   
   (f) $(\sqrt{3}+2 \sqrt{2})(\sqrt{3}+\sqrt{2})$
   
   
   
   

   $\begin{aligned} &\left( {\sqrt{3}+2\sqrt{2}} \right)\left( {\sqrt{3}+\sqrt{2}} \right)\\\\ =&\sqrt{3}\left( {\sqrt{3}+2\sqrt{2}} \right)+\sqrt{2}\left( {\sqrt{3}+2\sqrt{2}} \right)\\\\ =&3+2\sqrt{6}+\sqrt{6}+4\\\\ =&7+3\sqrt{6} \end{aligned}$

   
   
   (g) $(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})(\sqrt{x}+1)(\sqrt{x}-1)$
   
   
   
   

   $\begin{aligned} &\left( {\sqrt{7}-\sqrt{6}} \right)\left( {\sqrt{7}+\sqrt{6}} \right)\left( {\sqrt{x}+1} \right)\left( {\sqrt{x}-1} \right)\\\\ =&\left( {7-6} \right)\left( {x-1} \right)\\\\ =&x-1 \end{aligned}$

   
   
   (h) $\sqrt{75}-\dfrac{3}{4} \sqrt{48}-5 \sqrt{12}$
   
   
   
   

   $\begin{aligned} &\sqrt{{75}}-\frac{3}{4}\sqrt{{48}}-5\sqrt{{12}}\\\\ =&\sqrt{{25\times 3}}-\frac{3}{4}\sqrt{{16\times 3}}-5\sqrt{{4\times 3}}\\\\ =&5\sqrt{3}-\frac{3}{4}\times 4\sqrt{3}-5\times 2\sqrt{3}\\\\ =&5\sqrt{3}-3\sqrt{3}-10\sqrt{3}\\\\ =&-8\sqrt{3} \end{aligned}$

   
   
   (i) $\sqrt{2 x^{2}}+5 \sqrt{32 x^{2}}-2 \sqrt{98 x^{2}}$
   
   
   
   

   $\begin{aligned} &\sqrt{{2{{x}^{2}}}}+5\sqrt{{32{{x}^{2}}}}-2\sqrt{{98{{x}^{2}}}}\\\\ =&\sqrt{{2{{x}^{2}}}}+5\sqrt{{16\times 2{{x}^{2}}}}-2\sqrt{{49\times 2{{x}^{2}}}}\\\\ =&\left( {\sqrt{2}x} \right)+\left( {5\times 4\sqrt{2}x} \right)-\left( {2\times 7\sqrt{2}x} \right)\\\\ =&\sqrt{2}x+20\sqrt{2}x-14\sqrt{2}x\\\\ =&7\sqrt{2}x \end{aligned}$

   
   
   (j) $\sqrt{20 a^{3}}+a \sqrt{5 a}+\sqrt{80 a^{3}}$
   
   
   
   

   $\begin{aligned} &\sqrt{{20{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{80{{a}^{3}}}}\\\\ =&\sqrt{{4\times 5{{a}^{3}}}}+a\sqrt{{5a}}+\sqrt{{16\times 5{{a}^{3}}}}\\\\ =&2a\sqrt{{5a}}+a\sqrt{{5a}}+4a\sqrt{{5a}}\\\\ =&7a\sqrt{{5a}} \end{aligned}$

   
   
   
2. Rationalise the denominators and simplify.
   
   (a) $\dfrac{2}{\sqrt{5}}$
   
   
   
   

   $\begin{aligned} &\dfrac{2}{\sqrt{5}}\\\\ =&\dfrac{2}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\\\\ =&\dfrac{2\sqrt{5}}{5} \end{aligned}$

   
   
   (b) $\dfrac{5}{2+\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{5}{{2+\sqrt{3}}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{\left( {2+\sqrt{3}} \right)\left( {2-\sqrt{3}} \right)}}\\\\ =&\dfrac{{5\left( {2-\sqrt{3}} \right)}}{{4-3}}\\\\ =&5\left( {2-\sqrt{3}} \right) \end{aligned}$

   
   
   (c) $\dfrac{12}{\sqrt{5}-\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{12}}{{\sqrt{5}-\sqrt{3}}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{\left( {\sqrt{5}-\sqrt{3}} \right)\left( {\sqrt{5}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{{5-3}}\\\\ =&\dfrac{{12\left( {\sqrt{5}+\sqrt{3}} \right)}}{2}\\\\ =&6\left( {\sqrt{5}+\sqrt{3}} \right) \end{aligned}$

   
   
   (d) $\dfrac{\sqrt{2}+1}{2 \sqrt{2}-1}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{2}+1}}{{2\sqrt{2}-1}}\\\\ =&\dfrac{{\left( {\sqrt{2}+1} \right)\left( {2\sqrt{2}+1} \right)}}{{\left( {2\sqrt{2}-1} \right)\left( {2\sqrt{2}+1} \right)}}\\\\ =&\dfrac{{\sqrt{2}\left( {2\sqrt{2}+1} \right)+1\left( {2\sqrt{2}+1} \right)}}{{8-1}}\\\\ =&\dfrac{{4+\sqrt{2}+2\sqrt{2}+1}}{7}\\\\ =&\dfrac{{5+3\sqrt{2}}}{7} \end{aligned}$

   
   
   (e) $\dfrac{\sqrt{7}+3 \sqrt{2}}{\sqrt{7}-\sqrt{2}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{7}+3\sqrt{2}}}{{\sqrt{7}-\sqrt{2}}}\\\\ =&\dfrac{{\left( {\sqrt{7}+3\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}{{\left( {\sqrt{7}-\sqrt{2}} \right)\left( {\sqrt{7}+\sqrt{2}} \right)}}\\\\ =&\dfrac{{\sqrt{7}\left( {\sqrt{7}+\sqrt{2}} \right)+3\sqrt{2}\left( {\sqrt{7}+\sqrt{2}} \right)}}{{7-2}}\\\\ =&\dfrac{{7+\sqrt{{14}}+3\sqrt{{14}}+6}}{5}\\\\ =&\dfrac{{13+4\sqrt{{14}}}}{5} \end{aligned}$

   
   
   (f) $\dfrac{\sqrt{17}-\sqrt{11}}{\sqrt{17}+\sqrt{11}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{{17}}-\sqrt{{11}}}}{{\sqrt{{17}}+\sqrt{{11}}}}\\\\ =&\dfrac{{\left( {\sqrt{{17}}-\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}{{\left( {\sqrt{{17}}+\sqrt{{11}}} \right)\left( {\sqrt{{17}}-\sqrt{{11}}} \right)}}\\\\ =&\dfrac{{17-2\sqrt{{17}}\sqrt{{11}}+11}}{{17-11}}\\\\ =&\dfrac{{28-2\sqrt{{187}}}}{6}\\\\ =&\dfrac{{2\left( {14-\sqrt{{187}}} \right)}}{6}\\\\ =&\dfrac{{14-\sqrt{{187}}}}{3} \end{aligned}$

   
   
   (g) $\dfrac{1}{2 \sqrt{2}-\sqrt{3}}$
   
   
   
   

   $\begin{aligned} &\dfrac{1}{{2\sqrt{2}-\sqrt{3}}}\\\\ =&\dfrac{{1\cdot \left( {2\sqrt{2}+\sqrt{3}} \right)}}{{\left( {2\sqrt{2}-\sqrt{3}} \right)\left( {2\sqrt{2}+\sqrt{3}} \right)}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{{8-3}}\\\\ =&\dfrac{{2\sqrt{2}+\sqrt{3}}}{5} \end{aligned}$

   
   
   (h) $\dfrac{\sqrt{6}+1}{3-\sqrt{5}}$
   
   
   
   

   $\begin{aligned} &\dfrac{{\sqrt{6}+1}}{{3-\sqrt{5}}}\\\\ =&\dfrac{{\left( {\sqrt{6}+1} \right)\left( {3+\sqrt{5}} \right)}}{{\left( {3-\sqrt{5}} \right)\left( {3+\sqrt{5}} \right)}}\\\\ =&\dfrac{{\sqrt{6}\left( {3+\sqrt{5}} \right)+1\left( {3+\sqrt{5}} \right)}}{{9-5}}\\\\ =&\dfrac{{3\sqrt{6}+\sqrt{{30}}+3+\sqrt{5}}}{4}\\\\ =&\dfrac{{3+\sqrt{5}+3\sqrt{6}+\sqrt{{30}}}}{4} \end{aligned}$

   
   
   
3. Write as a single fraction.
   
   (a) $\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}$
   
   
   
   

   $\begin{aligned} & \dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1} \\\\ =& \dfrac{1}{\sqrt{3}+1} \times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-1} \times \dfrac{\sqrt{3}+1}{\sqrt{3}+1} \\\\ =& \dfrac{\sqrt{3}-1}{(\sqrt{3})^{2}-1^{2}}+\dfrac{\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}} \\\\ =& \dfrac{\sqrt{3}-1}{3-1}+\dfrac{\sqrt{3}+1}{3-1} \\\\ =& \dfrac{\sqrt{3}-1+\sqrt{3}+1}{2} \\\\ =& \dfrac{2 \sqrt{3}}{2} \\\\ =& \sqrt{3} \end{aligned}$

   
   
   (b) $\dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}}$
   
   
   
   

   $\begin{aligned} & \dfrac{2}{\sqrt{7}+\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \\\\ =& \dfrac{2}{\sqrt{7}+\sqrt{2}} \times \dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}-\sqrt{2}}+\dfrac{1}{\sqrt{7}-\sqrt{2}} \times \dfrac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\dfrac{\sqrt{7}+\sqrt{2}}{(\sqrt{7})^{2}-(\sqrt{2})^{2}} \\\\ =& \dfrac{2(\sqrt{7}-\sqrt{2})}{7-2}+\dfrac{\sqrt{7}+\sqrt{2}}{7-2} \\\\ =& \dfrac{2 \sqrt{7}-2 \sqrt{2}+\sqrt{7}+\sqrt{2}}{5} \\\\ =& \dfrac{3 \sqrt{7}-\sqrt{2}}{5} \end{aligned}$

   
   
   (c) $\dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}}$
   
   
   
   

   $\begin{aligned} & \dfrac{1}{3+\sqrt{3}}+\dfrac{1}{\sqrt{3}-3}+\dfrac{1}{\sqrt{3}} \\\\ =& \dfrac{3-3 \sqrt{3}}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{3 \sqrt{3}+3}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)}+\dfrac{-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{3-3 \sqrt{3}+3 \sqrt{3}+3-6}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& \dfrac{0}{\sqrt{3}(3+\sqrt{3})(\sqrt{3}-3)} \\\\ =& 0 \end{aligned}$

   
   
   (d) $\dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3}$
   
   
   
   

   $\begin{aligned} & \dfrac{7+\sqrt{5}}{7-\sqrt{5}}+\dfrac{\sqrt{11}-3}{\sqrt{11}+3} \\\\ =& \dfrac{(7+\sqrt{5})(7+\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}+\dfrac{(\sqrt{11}-3)(\sqrt{11}-3)}{(\sqrt{11}+3)(\sqrt{11}-3)} \\\\ =& \dfrac{54+14 \sqrt{5}}{49-5}+\dfrac{20-6 \sqrt{11}}{11-9} \\\\ =& \dfrac{54+14 \sqrt{5}}{44}+\dfrac{20-6 \sqrt{11}}{2} \\\\ =& \dfrac{27+7 \sqrt{5}}{22}+10-3 \sqrt{11} \\\\ =& \dfrac{27+7 \sqrt{5}+220-66 \sqrt{11}}{22} \\\\ =& \dfrac{247+7 \sqrt{5}-66 \sqrt{11}}{22} \end{aligned}$

   
   
   (e) $\dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}}$
   
   
   
   

   $\begin{aligned} & \dfrac{3+2 \sqrt{2}}{(\sqrt{3}-1)^{2}} \\\\ =& \dfrac{3+2 \sqrt{2}}{(\sqrt{3})^{2}-2 \sqrt{3+1}} \\\\ =& \dfrac{3+2 \sqrt{2}}{4-2 \sqrt{3}} \times \dfrac{4+2 \sqrt{3}}{4+2 \sqrt{3}} \\\\ =& \dfrac{(3+2 \sqrt{2})(4+2 \sqrt{3})}{4^{2}-(2 \sqrt{3})^{2}} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{16-12} \\\\ =& \dfrac{12+6 \sqrt{3}+8 \sqrt{2}+4 \sqrt{6}}{4} \\\\ =& \dfrac{6+3 \sqrt{3}+4 \sqrt{2}+2 \sqrt{6}}{2} \end{aligned}$

   
   
   (f) $\sqrt{\dfrac{x+1}{x-1}}+\sqrt{\dfrac{x-1}{x+1}}-\sqrt{\dfrac{1}{x^{2}-1}}$
   
   
   
   

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

   
   
   (g) $\sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}}$
   
   
   
   

   $\begin{aligned} & \sqrt{\dfrac{\sqrt[5]{32}+\sqrt{4}}{2^{-2}-2^{-3}}} \\\\ =& \sqrt{\dfrac{\sqrt[5]{2^{5}}+\sqrt{4}}{\dfrac{1}{2^{2}}-\dfrac{1}{2^{3}}}} \\\\ =& \sqrt{\dfrac{2+2}{\dfrac{1}{4}-\dfrac{1}{8}}} \\\\ =& \sqrt{\dfrac{4}{1 / 8}} \\\\ =& \sqrt{4 \times 8} \\\\ =& \sqrt{32} \\\\ =& 4 \sqrt{2} \end{aligned}$

   
   
   

Grade 10: Exercise (4.6) - Solution

A real valued function is one-to-one if every horizontal line intersects the graph of the function at most one point.

1. Determine whether each of the following function is a one-to-one function or not. If it is not one-to-one, explain why not.

(a) It is a one to one function.

(b) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(d) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(c) It is a one to one function.

2. Draw the graph of the each given function and determine whether each is a one-to-one function or not.
   
   (a) $f(x)=3 x+2$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & \ldots \\ \hline f(x) & \cdots & -4 & -1 & 2 & 5 & 8 & \ldots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (b) $f(x)=x-3$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & . \\ \hline f(x) & \cdots & -5 & -4 & -3 & 2 & 1 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (c) $f(x)=4 x^{2}$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 16 & 4 & 0 & 4 & 16 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (d) $f(x)=2|x|$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots- & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 4 & 2 & 0 & 2 & 4 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (e) $f(x)=\dfrac{2 x+3}{x+2}$
   
   $\begin{aligned} f(x) &=\dfrac{2 x+3}{x+2} \\\\ &=\dfrac{2 x+4-1}{(x+2)} \\\\ &=\dfrac{-1+2(x+2)}{(x+2)} \\\\ &=\dfrac{-1}{x+2}+2\\\\ &\text { horizontal arymptote: } y=2 \\\\ &\text { vertical asymptote : } x=-2 \\\\ &x=0, y=\frac{3}{2} \\\\ &y \text { -intercept }=\left(0, \frac{3}{2}\right) \\\\ &y=0, x=-\frac{3}{2} \\\\ &x \text { -intercept }=\left(-\frac{3}{2}, 0\right) \end{aligned}$
   
   

   It is a one to one function.

   (f) $f(x)=4 x^{2}\quad (0 \le x \le 4)$
   
   $\begin{array}{|c||c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 0 & 4 & 16 & 36 & 64 \\ \hline \end{array}$
   
   

   It is a one to one function.

   (g) $f(x)=\sqrt{x}\quad (x \ge 0)$
   
   $\begin{array}{|c||c|c|c|c|c|c|} \hline x & 0 & 1 & 4 & 9 & 25 & \ldots \\ \hline f(x) & 0 & 1 & 2 & 3 & 5 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.