Logarithms : Exercise (3.2) Solutions

1.           Write the following equations in logarithmic form.

              (a)   $3^{4}=81$

              (b)   $9^{\frac{3}{2}}=27$

              (c)   $10^{-3}=0.001$

              (d)   $3^{-1}=\displaystyle\frac{1}{3}$

              (e)   $\left(\displaystyle\frac{1}{4}\right)^{-3}=64$

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$\begin{array}{l} \text{(a)}\ \ {{3}^{4}}=81\\ \ \ \ \ \ {{\log }_{3}}81=4\\\\ \text{(b)}\ \ {{9}^{{\frac{3}{2}}}}=27\\ \ \ \ \ \ {{\log }_{9}}27=\displaystyle\frac{3}{2}\\\\ \text{(c)}\ \ {{10}^{{-3}}}=0.001\\ \ \ \ \ \ {{\log }_{{10}}}0.001=-3\\\\ \text{(d)}\ \ {{3}^{{-1}}}=\displaystyle\frac{1}{3}\\ \ \ \ \ \ {{\log }_{3}}\displaystyle\frac{1}{3}=-1\\\\ \text{(e)}\ \ {{\left( {\displaystyle\frac{1}{4}} \right)}^{{-3}}}=64\\ \ \ \ \ \ {{\log }_{{\frac{1}{4}}}}64=-3 \end{array}$

2.           Write the following equations in exponential form.

             (a)   $\log _{10} 3=0.4771$

             (b)   $\log _{6} 0.001=-3.855$

             (c)   $\log _{144} 12=\displaystyle\frac{1}{2}$

             (d)   $-5=\log _{3}\displaystyle \frac{1}{243}$

             (e)   $\log _{x} .7=y^{2},$ where, $0<x<1$

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$\begin{array}{l} \text{(a)}\ \ {{\log }_{{10}}}3=0.4771\\ \ \ \ \ \ {{3}^{{0.4771}}}=10\\\\ \text{(b)}\ \ {{\log }_{6}}0.001=-3.855\\ \ \ \ \ \ {{6}^{{-3.855}}}=0.001\\\\ \text{(c)}\ \ {{\log }_{{144}}}12=\displaystyle \frac{1}{2}\\ \ \ \ \ \ {{144}^{{\frac{1}{2}}}}=12\\\\ \text{(d)}\ \ -5={{\log }_{3}}\frac{1}{{243}}\\ \ \ \ \ \ \displaystyle \frac{1}{{243}}={{3}^{{-5}}}\\\\ \text{(e)}\ \ {{\log }_{x}}.7={{y}^{2}},\ \ \text{where}\ 0<x<1\\ \ \ \ \ {{x}^{{{{y}^{2}}}}}=.7 \end{array}$

3.           Solve the following equations

             (a)   $\log _{7} 49=x$

             (b)   $\log _{x} 10=1$

             (c)   $\log _{\sqrt{3}} x=2$

             (d)   $x^{\log _{x} x}=5$

             (e)   $\log _{0.2} 5=x$

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$\begin{array}{l} \text{(a)}\ \ {{\log }_{7}}49=x\\ \ \ \ \ \ {{7}^{x}}=49\\ \ \ \ \ \ {{7}^{x}}={{7}^{2}}\\ \ \ \ \ \ x=2\\\\ \text{(b)}\ \ {{\log }_{x}}10=1\\ \ \ \ \ \ {{x}^{1}}=10\\ \ \ \ \ \ x=10\\\\ \text{(c)}\ \ {{\log }_{{\sqrt{3}}}}x=2\ \\ \ \ \ \ x={{\left( {\sqrt{3}} \right)}^{2}}\\ \ \ \ \ \ x=3\\\\ \text{(d)}\ \ {{x}^{{{{{\log }}_{x}}x}}}=5\\ \ \ \ \ \ x=5\\\\ \text{(e)}\ \ {{\log }_{{0.2}}}5=x\\ \ \ \ \ \ {{0.2}^{x}}=5\\ \ \ \ \ \ {{\left( {\displaystyle\frac{1}{5}} \right)}^{x}}=5\\ \ \ \ \ \ {{5}^{{-x}}}={{5}^{1}}\\\ \ \ \ \ x=-1\ \ \end{array}$

4.           Evaluate.

             (a)   $9^{\log _{9} 2}+3^{\log _{3} 8}$

             (b)   $\log _{4} 4^{5} \times \log _{10} 10^{2}$

             (c)   $7^{\log _{7} 9}+\log _{2}\left(\displaystyle\frac{1}{2}\right)$

             (d)   $\log _{\frac{1}{2}} \displaystyle\frac{1}{8}-4 \log _{10} 10$

             (e)   $10^{1-\log _{10} 3}$

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$\begin{array}{l} \text{(a)}\ \ \ \ {{9}^{{{{{\log }}_{9}}2}}}+{{3}^{{{{{\log }}_{3}}8}}}\\ \ \ \ \ \ =2+8\\\ \ \ \ \ =10\\\\ \text{(b)}\ \ \ \ {{\log }_{4}}{{4}^{5}}\times {{\log }_{{10}}}{{10}^{2}}\\ \ \ \ \ \ =5\times 2\\\ \ \ \ \ =10\\\\ \text{(c)}\ \ \ \ \ {{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}\left( {\displaystyle\frac{1}{2}} \right)\\ \ \ \ \ \ ={{7}^{{{{{\log }}_{7}}9}}}+{{\log }_{2}}{{2}^{{-1}}}\\ \ \ \ \ \ =9-1\\\ \ \ \ \ =8\\\\ \text{(d)}\ \ \ \ \ {{\log }_{{\frac{1}{2}}}}\displaystyle\frac{1}{8}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ ={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{3}}-4\cdot {{\log }_{{10}}}10\\ \ \ \ \ \ =3-4\cdot (1)\\\ \ \ \ \ =-1\\\\ \text{(e)}\ \ \ \ \ {{10}^{{1-{{{\log }}_{{10}}}3}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{{{{{10}}^{{{{{\log }}_{{10}}}3}}}}}\\ \ \ \ \ \ =\displaystyle\frac{{10}}{3} \end{array}$

5.           Find the value of $x$ in each of the following problems.

             (a)   $\log _{3}(2 x-5)=2,$ where $x>\displaystyle\frac{5}{2}$

             (b)   $\log _{77}\left(\log _{7} x\right)=0,$ where $x>0$

             (c)   $8+3^{x}=10,$ given that $\log _{3} 2=0.6309$

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$\begin{array}{l}\text{(a)}\ \ {{\log }_{3}}(2x-5)=2,\text{where}\ x>\displaystyle \frac{5}{2}\\\ \ \ \ \ 2x-5={{3}^{2}}\\\ \ \ \ \ 2x-5=9\\\ \ \ \ \ x=7\\\\\text{(b)}\ \ {{\log }_{{77}}}\left( {{{{\log }}_{7}}x} \right)=0,\ \ \text{where}\ x>0\\\ \ \ \ \ {{\log }_{7}}x={{77}^{0}}\\\ \ \ \ \ {{\log }_{7}}x=1\\\ \ \ \ \ x={{7}^{1}}=7\\\\\text{(c)}\ \ 8+{{3}^{x}}=10,\ \text{given that}\ {{\log }_{3}}2=0.6309\\\ \ \ \ \ {{3}^{x}}=2\\\ \ \ \ \ x={{\log }_{3}}2\\\ \ \ \ \ x=0.6309\end{array}$

Exercise (5.5) : Solving Linear Quadratic Systems Algebraically

1.           Find the solution set of each of the systems of equations:

              (a)   $x^2-y^2=9$

                     $x+y=1$

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$\begin{array}{l}{{x}^{2}}-{{y}^{2}}=9\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=1\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=1-x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting }y=1-x\text{ in equation }(1)\text{,}\\\\{{x}^{2}}-{{\left( {1-x} \right)}^{2}}=9\\\\{{x}^{2}}-\left( {1-2x+{{x}^{2}}} \right)=9\\\\{{x}^{2}}-1+2x-{{x}^{2}}=9\\\\2x=10\\\\x=5\\\\\text{Substituting }x=5\text{ in equation }(3),\\\\y=1-5=-4\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {5,-4} \right)} \right\}\text{ }\end{array}$

              (b)   $y=\displaystyle \frac{8}{x}$

                     $y=7+x$

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$\begin{array}{l}y=\displaystyle \frac{8}{x}\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\y=7+x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\\displaystyle \frac{8}{x}=7+x\\\\\therefore \ {{x}^{2}}+7x=8\\\\{{x}^{2}}+7x-8=0\\\\(x+8)(x-1)=0\\\\x=-8\ \text{or}\ x=1\\\\\text{When}\ x=-8,\ y=7-8=-1\\\\\text{When}\ x=1,\ y=7+1=8\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-8,-1} \right),\left( {1,8} \right)} \right\}\text{ }\end{array}$

              (c)   $x^2+5x+y=4$

                     $x+y=8$

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$\begin{array}{l}{{x}^{2}}+5x+y=4\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=8\ \ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=8-x\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=8-x\text{ in equation }(1),\\\\{{x}^{2}}+5x+8-x=4\\\\{{x}^{2}}+4x+4=0\\\\{{(x+2)}^{2}}=0\\\\x=-2\\\\\text{Substituting}\ x=-2\text{ in equation }(3),\\\\y=8-(-2)=10\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-2,10} \right)} \right\}\text{ }\end{array}$

2.           The sum of squares of two numbers is 58. If the first number and twice the second add up to 13, find the numbers.

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$\begin{array}{l}\text{Let}\ \text{the two numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\{{x}^{2}}+{{y}^{2}}=58\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\x+2y=13\ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\x=13-2y\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ x=13-2y\text{ in equation }(1),\\\\{{\left( {13-2y} \right)}^{2}}+{{y}^{2}}=58\\\\169-52y+4{{y}^{2}}+{{y}^{2}}=58\\\\169-52y+5{{y}^{2}}=58\\\\\therefore \ \ 5{{y}^{2}}-52y+111=0\\\\(5y-37)(y-3)=0\\\\y=\displaystyle \frac{{37}}{5}\ \text{or}\ y=3\\\\\text{When}\ y=\displaystyle \frac{{37}}{5},x=13-2\left( {\displaystyle \frac{{37}}{5}} \right)\text{=}-\displaystyle \frac{9}{5}\text{ }\\\\\text{When}\ y=3,x=13-2\left( 3 \right)\text{=}7\text{ }\\\\\therefore \ \ \text{The two numbers are }-\displaystyle \frac{9}{5}\ \text{and }\displaystyle \frac{{37}}{5}\ \text{or 7 and 3}\text{.}\end{array}$

3.           The sum of the reciprocals of two positive numbers is $\displaystyle \frac{7}{36}$ and the product of the numbers is 108. Find the numbers.

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$\begin{array}{l}\text{Let}\ \text{the two positive numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}=\displaystyle \frac{7}{{36}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\xy=108\ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=\displaystyle \frac{{108}}{x}\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=\displaystyle \frac{{108}}{x}\text{ in equation }(1),\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{{\displaystyle \frac{{108}}{x}}}=\displaystyle \frac{7}{{36}}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{x}{{108}}=\displaystyle \frac{7}{{36}}\\\\\text{Multiplying both sides of equation by}\ 108x,\\\\108+{{x}^{2}}=21x\\\\\therefore \ \ {{x}^{2}}+21x-108=0\\\\(x-9)(x-12)=0\\\\x=9\ \text{or}\ x=12\\\\\text{When}\ x=9,y=\displaystyle \frac{{108}}{9}=12\\\\\text{When}\ x=12,y=\displaystyle \frac{{108}}{{12}}=9\\\\\therefore \ \ \text{The two positive numbers are }9\text{ and 12}\text{.}\end{array}$

QUADRATIC FUNCTIONS : EXERCISE (5.4) SOLUTIONS

            Solve the following equations by using the quadratic formula.

1.          $x^{2}+2 x-1=0$

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$\begin{array}{l} \ \ \ \ {{x}^{2}}+2x-1=0\\\\ \ \ \ \ \text{Comparing with}\ a{{x}^{2}}+bx+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=2,\ c=-1\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{2}^{2}}-4(2)(-1)=8>0\\\\ \ \ \ \ \therefore \ \text{There are two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm \sqrt{{{{2}^{2}}-4(2)(-1)}}}}{{2(1)}}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm \sqrt{8}}}{2}\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-2\pm 2\sqrt{2}}}{2}\\\\ \ \ \ \ \ \ \ \ \ =-1\pm \sqrt{2}\\\\ \ \ \ \ \therefore \ \ x=-1-\sqrt{2}\ \text{or}\ x=-1+\sqrt{2}\end{array}$

2.          $x^{2}+4 x+4=0$

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$ \begin{array}{l}\ \ \ {{x}^{2}}+4x+4=0\\\\\;\;\;\;\text{Comparing with}\;a{{x}^{2}}+bx+c=0,\;\text{we get}\;\\\;\;\;\\\ \ \ a=1,\;b=4,\;c=4\;\;\\\;\;\\\ \ \ \text{discriminant}={{b}^{2}}-4ac={{4}^{2}}-4(1)(4)=0\;\\\;\;\;\\\ \ \therefore \;\text{There is one real solution for }x.\;\;\;\\\;\\\ \ \therefore \;\;x=-\displaystyle \frac{b}{{2a}}=-\displaystyle \frac{4}{{2(1)}}=-2\end{array}$

3.          $p^{2}-6 p+3=0$

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$ \begin{array}{l}\ \ \ \ {{p}^{2}}-6p+3=0\\\\ \ \ \ \ \text{Comparing with}\ a{{p}^{2}}+bp+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=-6,\ c=3\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-6)}^{2}}-4(1)(3)=24>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{6\pm \sqrt{{24}}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{6\pm 2\sqrt{6}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =3\pm \sqrt{6}\\\\ \ \ \ \ \therefore \ \ x=3-\sqrt{6}\ \text{or}\ x=3+\sqrt{6}\end{array}$

4.          $t^{2}-4 t-8=0$

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$\begin{array}{l} \ \ \ \ {{t}^{2}}-4t-8=0\\\\ \ \ \ \ \text{Comparing with}\ a{{t}^{2}}+bt+c=0,\ \text{we get}\\\\ \ \ \ \ a=1,\ b=-4,\ c=-8\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-4)}^{2}}-4(1)(-8)=48>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4\pm \sqrt{{48}}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4\pm 4\sqrt{3}}}{2}\\\\ \ \ \ \ \ \ \ \ \ \ =2\pm 2\sqrt{3}\\\\ \ \ \ \ \therefore \ \ x=2-2\sqrt{3}\ \text{or}\ x=2+2\sqrt{3} \end{array}$

5.          $3 q^{2}-12 q+11=0$

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$\begin{array}{l}\ \ \ \ 3{{q}^{2}}-12q+11=0\\\\ \ \ \ \ \text{Comparing with}\ a{{q}^{2}}+bq+c=0,\ \text{we get}\\\\ \ \ \ \ a=3,\ b=-12,\ c=11\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{(-12)}^{2}}-4(3)(11)=12>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{12\pm \sqrt{{12}}}}{6}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{12\pm 3\sqrt{3}}}{6}\\\\ \ \ \ \ \ \ \ \ \ \ =2\pm \displaystyle \frac{{\sqrt{3}}}{2}\\\\ \ \ \ \ \therefore \ \ x=2-\displaystyle \frac{{\sqrt{3}}}{2}\ \text{or}\ x=2+\displaystyle \frac{{\sqrt{3}}}{2} \end{array}$

6.          $5 z^{2}+3 z-4=0$

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$\begin{array}{l} \ \ \ \ 5{{z}^{2}}+3z-4=0\\\\ \ \ \ \ \text{Comparing with}\ a{{z}^{2}}+bz+c=0,\ \text{we get}\\\\ \ \ \ \ a=5,\ b=3,\ c=-4\\\\ \ \ \ \ \text{discriminant}={{b}^{2}}-4ac={{3}^{2}}-4(5)(-4)=89>0\\\\ \ \ \ \ \therefore \ \text{There are}\ \text{two real solutions for }x.\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\\\\ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-3\pm \sqrt{{89}}}}{{10}}\\\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{{-3-\sqrt{{89}}}}{{10}}\ \text{or}\ x=\displaystyle \frac{{-3+\sqrt{{89}}}}{{10}} \end{array}$

QUADRATIC FUNCTIONS : EXERCISE (5.3) SOLUTIONS

1.         Find the discriminant of each of the following quadratic functions. Also find the number of $x$ -intercepts of each of the functions.

           $\begin{array}{l} \text{(a)}\ \ y=3 x^{2}-4 x+3\\\\ \text{(b)}\ \ y=2 x^{2}-4 x-3\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=-x^{2}+6 x-9\\\\ \text{(e)}\ \ y=-3 x^{2}-12 x-7\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$

           $\text{(a)}\ \ y=3 x^{2}-4 x+3$

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$\begin{array}{l} \ \ \ \ \ y=3{{x}^{2}}-4x+3\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=3,\ b=-4,\ c=3\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-4)}^{2}}-4(3)(3)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-20<0\\\\ \ \ \ \ \ \therefore \ \ \text{There is no }x-\text{intercept point.} \end{array}$

           $\text{(b)}\ \ y=2 x^{2}-4 x-3$

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$\begin{array}{l} \ \ \ \ \ y=2{{x}^{2}}-4x-3\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=2,\ b=-4,\ c=-3\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-4)}^{2}}-4(2)(-3)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

           $\text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4$

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$\begin{array}{l}\ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=\displaystyle \frac{1}{2},\ b=1,\ c=-4\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(1)}^{2}}-4\left( {\displaystyle \frac{1}{2}} \right)(-4)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

           $\text{(d)}\ \ y=-x^{2}+6 x-9$

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$\begin{array}{l} \ \ \ \ \ y=-{{x}^{2}}+6x-9\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-1,\ b=6,\ c=-9\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(6)}^{2}}-4(-1)(-9)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\ \ \ \ \ \ \therefore \ \ \text{There is one }x-\text{intercept point.}\ \end{array}$

           $\text{(e)}\ \ y=-3 x^{2}-12 x-7$

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$ \begin{array}{l} \ \ \ \ \ y=-3{{x}^{2}}-12x-7\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-3,\ b=-12,\ c=-7\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-12)}^{2}}-4\left( {-3} \right)(-7)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =60>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

           $\text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4$

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$\begin{array}{l} \ \ \ \ \ y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4\\\\ \ \ \ \ \ \text{Comparing with }a{{x}^{2}}+bx+c\text{ we have,}\\\\ \ \ \ \ \ a=-\displaystyle \frac{1}{2},\ b=-3,\ c=-4\\\\ \ \ \ \ \ \therefore \ \ \text{Discriminant}\ \ ={{b}^{2}}-4ac\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(-3)}^{2}}-4\left( {-\displaystyle frac{1}{2}} \right)(-4)\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1>0\\\\ \ \ \ \ \ \therefore \ \ \text{There are two }x-\text{intercept points.} \end{array}$

2.         Find the intercept form of each of the quadratic functions. Also find the $y$ -intercept, axis of symmetry, vertex, and range of each of the functions.

           $\begin{array}{l} \text{(a)}\ \ y=2 x^{2}-2 x-12\\\\ \text{(b)}\ \ y=3 x^{2}-6 x+3\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=2 x^{2}-5 x-3\\\\ \text{(e)}\ \ y=-6 x^{2}-7 x+5\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$

           $\text{(a)}\ \ y=2 x^{2}-2 x-12$

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$ \begin{array}{l} \ \ \ \ y=2{{x}^{2}}-2x-12\\\\ \ \ \ \ y=2({{x}^{2}}-x-6)\\\\ \ \ \ \ y=2(x+2)(x-3)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-12\\\\ \ \ \ \ \therefore \ y-\text{intercept point}=(0,\ -12)\\\\ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=2,\ p=-2,\ q=3\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{p+q}}{2}\Rightarrow x=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{1}{2},\ y=2{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)-12=-\displaystyle \frac{{25}}{2}\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {\displaystyle \frac{1}{2},-\displaystyle \frac{{25}}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{{25}}{2}} \right\} \end{array}$

           $\text{(b)}\ \ y=3 x^{2}-6 x+3$

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$\begin{array}{l} \ \ \ \ y=3{{x}^{2}}-6x+3\;\\\\ \ \ \ \ y=3({{x}^{2}}-2x+1)\\\\ \ \ \ \ y=3(x-1)(x-1)\\\\ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ 3)\\\\ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=3,\ p=1,\ q=1\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{1+1}}{2}\Rightarrow x=1\\\\ \ \ \ \ \text{When}\ x=1,\ y=3(1-1)(1-1)=0\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {1,0} \right) \end{array}$

           $\text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4$

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$\begin{array}{l} \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ y=\displaystyle \frac{1}{2}({{x}^{2}}+2x-8)\\\\ \ \ \ \ y=\displaystyle \frac{1}{2}(x+4)(x-2)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -4)\\\\ \text{ }\ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=\displaystyle \frac{1}{2},\ p=-4,\ q=2\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-4+2}}{2}\Rightarrow x=-1\\\\ \ \ \ \ \text{When}\ x=-1,\ y=\displaystyle \frac{1}{2}(-1+4)(-1-2)=-\displaystyle \frac{9}{2}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-1,-\displaystyle \frac{9}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{9}{2}} \right\} \end{array}$

           $\text{(d)}\ \ y=2 x^{2}-5 x-3$

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$\begin{array}{l} \ \ \ \ y=2{{x}^{2}}-5x-3\\\\ \ \ \ \ y=2\left( {{{x}^{2}}-\displaystyle \frac{5}{2}x-\displaystyle \frac{3}{2}} \right)\\\\ \ \ \ \ y=2\left( {x+\displaystyle \frac{1}{2}} \right)\left( {x-3} \right)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-3\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -3)\\\\ \text{ }\ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=2,\ p=-\displaystyle \frac{1}{2},\ q=3\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-\displaystyle \frac{1}{2}+3}}{2}\Rightarrow x=\displaystyle \frac{5}{4}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=2\left( {\displaystyle \frac{5}{4}+\displaystyle \frac{1}{2}} \right)\left( {\displaystyle \frac{5}{4}-3} \right)=-\displaystyle \frac{{49}}{8}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {\displaystyle \frac{5}{4},-\displaystyle \frac{{49}}{8}} \right)\\\\\ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\ge -\displaystyle \frac{{49}}{8}} \right\} \end{array}$

           $\text{(e)}\ \ y=-6 x^{2}-7 x+5$

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$\begin{array}{l}\ \ \ \ y=-6{{x}^{2}}-7x+5\\\\ \ \ \ \ y=-\left( {6{{x}^{2}}-7x-5} \right)\\\\ \ \ \ \ y=-(3x+5)(2x-1)\\\\\ \ \ \ y=-3\left( {x+\displaystyle \frac{5}{3}} \right)2\left( {x-\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ y=-6\left( {x+\displaystyle \frac{5}{3}} \right)\left( {x-\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ \text{When}\ x=0,\ y=5\\\\\ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ 5)\\\\ \ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=-6,\ p=-\displaystyle \frac{5}{2},\ q=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-\displaystyle \frac{5}{3}+\displaystyle \frac{1}{2}}}{2}\Rightarrow x=-\displaystyle \frac{7}{{12}}\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=-6\left( {-\displaystyle \frac{7}{{12}}+\displaystyle \frac{5}{3}} \right)\left( {-\displaystyle \frac{7}{{12}}-\displaystyle \frac{1}{2}} \right)=\displaystyle \frac{{169}}{{24}}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-\displaystyle \frac{7}{{12}},\displaystyle \frac{{169}}{{24}}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\le \displaystyle \frac{{169}}{{24}}} \right\} \end{array}$

           $ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 $

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$\begin{array}{l} \ \ \ \ y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4\\\\ \ \ \ \ y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+6x+8} \right)\\\\ \ \ \ \ y=-\displaystyle \frac{1}{2}(x+4)(x+2)\\\\ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \therefore \ y-\text{intercept}\ \text{point}\ =(0,\ -4)\\\\ \ \ \ \ \text{Comparing with}\ y=a(x-p)(x-q),\\\\ \ \ \ \ a=-\displaystyle \frac{1}{2},\ p=-4,\ q=-2\\\\ \ \ \ \ \therefore \text{Axis}\ \text{of symmetry}\ \text{: }x=\displaystyle \frac{{-4-2}}{2}\Rightarrow x=-3\\\\ \ \ \ \ \text{When}\ x=\displaystyle \frac{5}{4},\ y=-\displaystyle \frac{1}{2}(-3+4)(-3+2)=\displaystyle \frac{1}{2}\;\\\\ \ \ \ \ \therefore \text{vertex}\ =\left( {-3,\displaystyle \frac{1}{2}} \right)\\\\ \ \ \ \ \therefore \text{range}\ =\left\{ {y\ |\ y\le \displaystyle \frac{1}{2}} \right\} \end{array}$

Quadratic Functions : Exercise (5.2) Solutions

1.         Find the vertex form of each of the following quadratic functions. Find also $y$ -intercept, axis of symmetry, vertex, and range of each of the functions.

           $\begin{array}{l} \text{(a)}\ \ y=2 x^{2}+4 x+3\\\\ \text{(b)}\ \ y=3 x^{2}-6 x+2\\\\ \text{(c)}\ \ y=\displaystyle\frac{1}{2} x^{2}+x-4\\\\ \text{(d)}\ \ y=-2 x^{2}+2 x+3\\\\ \text{(e)}\ \ y=-3 x^{2}-12 x-7\\\\ \text{(f)}\ \ y=-\displaystyle\frac{1}{2} x^{2}-3 x-4 \end{array}$


           $ \text{(a)}\;\;y=2{{x}^{2}}+4x+3$

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$ \begin{array}{l} \ \ \ \ \ y=2{{x}^{2}}+4x+3\\\\ \ \ \ \ \ y=2\left( {{{x}^{2}}+2x+1} \right)+1\\\\ \ \ \ \ \ y=2{{\left( {x+1} \right)}^{2}}+1\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 3).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=2,\ h=-1,\ k\ =1\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry:} x=h\Rightarrow x=-1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(-1,1)\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ 1\} \end{array}$

           $\text{(b)}\;\;y=3{{x}^{2}}-6x+2$

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$\begin{array}{l}\ \ \ \ \ y=3{{x}^{2}}-6x+3-1\\\\ \ \ \ \ \ y=3\left( {{{x}^{2}}-2x+1} \right)-1\\\\ \ \ \ \ \ y=3{{\left( {x-1} \right)}^{2}}-1\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=2\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 2).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=3,\ h=1,\ k\ =-1\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(1,-1)\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ -1\}\end{array}$

           ${\text{(c)}\;\;y=\displaystyle\frac{1}{2}{{x}^{2}}+x-4}$

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$\begin{array}{l} \ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+x-4\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}{{x}^{2}}+\displaystyle \frac{2}{2}x+\displaystyle \frac{1}{2}-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}\left( {{{x}^{2}}+2x+1} \right)-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ y=\displaystyle \frac{1}{2}{{\left( {x+1} \right)}^{2}}-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=-4\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ -4).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=\displaystyle \frac{1}{2},\ h=-1,\ k\ =-\displaystyle \frac{9}{2}\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=-1\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(-1,-\displaystyle \frac{9}{2})\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \ge \ k\}=\{y\ |\ y\ \ge \ -\displaystyle \frac{9}{2}\} \end{array}$

           $ \text{(d)}\;\ y=-2{{x}^{2}}+2x+3$

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$\begin{array}{l} \ \ \ \ \ y=-2{{x}^{2}}+2x+3\\\\ \ \ \ \ \ y=-2\left( {{{x}^{2}}-x} \right)+3\\\\ \ \ \ \ \ y=-2\left( {{{x}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)x+\displaystyle \frac{1}{4}} \right)+3+\displaystyle \frac{1}{2}\\\\ \ \ \ \ \ y=-2{{\left( {x-\displaystyle \frac{1}{2}} \right)}^{2}}+\displaystyle \frac{7}{2}\\\\ \ \ \ \ \ \text{When}\ x=0,\ y=3\\\\ \ \ \ \ \ \therefore \ y-\text{intercept}\ (0,\ 3).\\\\ \ \ \ \ \ \text{Comparing with }a{{\left( {x-h} \right)}^{2}}+k,\\\\ \ \ \ \ \ a=-2,\ h=\displaystyle \frac{1}{2},\ k\ =\displaystyle \frac{7}{2}\\\\ \ \ \ \ \ \therefore \ \text{axis of symmetry: }x=h\Rightarrow x=\displaystyle \frac{1}{2}\\\\ \ \ \ \ \ \ \ \text{vertex}=(h,k)=(\displaystyle \frac{1}{2},\displaystyle \frac{7}{2})\\\\ \ \ \ \ \ \ \ \text{range}=\{y\ |\ y\ \le \ k\}=\{y\ |\ y\ \le \ \displaystyle \frac{7}{2}\} \end{array}$

           ${\text{(e)}\;\;y=-3{{x}^{2}}-12x-7}$

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$\begin{array}{*{20}{l}} {\;\;\;\;\;y=-3{{x}^{2}}-12x-7} \\\\ {\;\;\;\;\;y=-3\left( {{{x}^{2}}+4x} \right)-7}\\\\ {\;\;\;\;\;y=-3\left( {{{x}^{2}}+4x+4} \right)-7+12}\\\\ {\;\;\;\;\;y=-3{{{\left( {x+2} \right)}}^{2}}+5} \\\\ {\;\;\;\;\;\text{When}\;x=0,\;y=-7}\\\\ {\;\;\;\;\;\therefore \;y-\text{intercept}\;(0,\;-7).}\\\\ {\;\;\;\;\;\text{Comparing with }a{{{\left( {x-h} \right)}}^{2}}+k,}\\\\ {\;\;\;\;\;a=-3,\;h=-2,\;k\;=5} \\ {} \\ {\;\;\;\;\;\therefore \;\text{axis of symmetry: }x=h\Rightarrow x=-2} \\\\ {\;\;\;\;\;\;\;\text{vertex}=(h,k)=(-2,5)}\\\\ {\;\;\;\;\;\;\;\text{range}=\{y\;|\;y\;\le \;k\}=\{y\;|\;y\;\le \;5\}} \end{array}$

           ${\text{(f)}\;\;y=-\displaystyle\frac{1}{2}{{x}^{2}}-3x-4}$

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$\begin{array}{*{20}{l}} {\;\;\;\;\;y=-\displaystyle \frac{1}{2}{{x}^{2}}-3x-4} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+6x} \right)-4} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}\left( {{{x}^{2}}+2(3)x+9} \right)-4+\displaystyle \frac{9}{2}} \\ {} \\ {\;\;\;\;\;y=-\displaystyle \frac{1}{2}{{{\left( {x+3} \right)}}^{2}}+\displaystyle \frac{1}{2}} \\ {} \\ {\;\;\;\;\;\text{When}\;x=0,\;y=-4} \\ {} \\ {\;\;\;\;\;\therefore \;y-\text{intercept}\;(0,\;-4).} \\ {} \\ {\;\;\;\;\;\text{Comparing with }a{{{\left( {x-h} \right)}}^{2}}+k,} \\ {} \\ {\;\;\;\;\;a=-\displaystyle \frac{1}{2},\;h=-3,\;k\;=\displaystyle \frac{1}{2}} \\ {} \\ {\;\;\;\;\;\therefore \;\text{axis of symmetry: }x=h\Rightarrow x=-3} \\ {} \\ {\;\;\;\;\;\;\;\text{vertex}=(h,k)=(-3,\displaystyle \frac{1}{2})} \\ {} \\ {\;\;\;\;\;\;\;\text{range}=\{y\;|\;y\;\le \;k\}=\{y\;|\;y\;\le \;\displaystyle \frac{1}{2}\}} \end{array}$

2.         Find two positive numbers whose sum is $20$ and whose product is as large as possible.

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Let the two positive numbers be $x$ and $y$.

By the problem,

$x + y = 20$

$\therefore\ y=20-x$

Let the product of the numbers be $p$.

Hence,

$\begin{aligned} p &=xy\\\\ &=x(20-x)\\\\ &=20x-x^2\\\\ &=-(x^2-20x)\\\\ &=-(x^2-20x + 100 )+100\\\\ &=100-(x-10)^2\\\\ \therefore\ p&\le100 \end{aligned}$

Thus, maximum value of $p$ is $100$ when $(x-10)^2 = 0$.

$\therefore\ x = 10$.

When $x=10, y = 20 - x = 20 - 10 = 10$.

Therefore, the twopositive numbers are $x=10$ and $y=10$.

3.         What is the largest area possible for a rectangle whose perimeter is $16$ cm?

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Let the lenght and breadth of the rectangle be $x$ and $y$ respectively.

By the problem,

$2x + 2y = 16$

$\therefore\ y=8-x$

Let the area of the rectangle be $A$.

Hence,

$\begin{aligned} A &=xy\\\\ &=x(8-x)\\\\ &=8x-x^2\\\\ &=-(x^2-8x)\\\\ &=-(x^2-2\cdot 4\cdot x + 16 )+16\\\\ &=16-(x-4)^2\\\\ \therefore\ A&\le 16 \end{aligned}$

Thus, the maximum area $(A)$ of the rectangle is $16$ cm².

EXPONENTS AND RADICALS : EXERCISE (2.2) SOLUTIONS

1.        Evaluate the following.

         (a)     $(125)^{\frac{2}{3}}$
         (b)     $(81)^{-\frac{3}{2}}$
         (c)     $(-27)^{\frac{2}{3}}$
         (d)     $\left(\displaystyle\frac{16}{81}\right)^{-\frac{3}{4}}$
         (e)     $\left(\displaystyle\frac{-125}{8} \div \frac{1}{64}\right)^{\frac{1}{3}}$
         (f)     $(0.125)^{-\frac{2}{3}}$
         (g)     $\left(\displaystyle\frac{64}{27}\right)^{-\frac{2}{3}}$
         (h)     $(-4)^{-1}+(-1)^{-4}$

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$ \displaystyle \begin{array}{l} \text{(a)}\ \ {{(125)}^{{\frac{2}{3}}}}={{({{5}^{3}})}^{{\frac{2}{3}}}}={{5}^{2}}=25\\\\ \text{(b)}\ \ {{(81)}^{{-\frac{3}{2}}}}={{({{9}^{2}})}^{{-\frac{3}{2}}}}={{9}^{{-3}}}=\displaystyle\frac{1}{{{{9}^{3}}}}=\displaystyle\frac{1}{{729}}\\\\ \text{(c)}\ \ {{(-27)}^{{\frac{2}{3}}}}={{\left( {{{{(-3)}}^{3}}} \right)}^{{\frac{2}{3}}}}={{(-3)}^{2}}=9\\\\ \text{(d)}\ \ {{\left( {\displaystyle\frac{{16}}{{81}}} \right)}^{{-\frac{3}{4}}}}={{\left( {{{{\left( {\frac{2}{3}} \right)}}^{4}}} \right)}^{{-\frac{3}{4}}}}={{\displaystyle\left( {\frac{2}{3}} \right)}^{{-3}}}={{\displaystyle\left( {\frac{3}{2}} \right)}^{3}}=\displaystyle\frac{{27}}{8}\\\\ \text{(e)}\ \ \ {{\left( {\displaystyle\frac{{-125}}{8}\div \frac{1}{{64}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {\displaystyle\frac{{-125}}{8}\times 64} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {-125\times 8} \right)}^{{\frac{1}{3}}}}\\\ \ \ ={{\left( {{{{\left( {-5} \right)}}^{3}}\times {{2}^{2}}} \right)}^{{\frac{1}{3}}}}\\\ \ \ =-5\times 2\\\ \ \ =-10\\ \text{(f)}\ \ {{(0.125)}^{{-\frac{2}{3}}}}={{\left( {{{{0.5}}^{3}}} \right)}^{{-\frac{2}{3}}}}={{\left( {0.5} \right)}^{{-2}}}={{\left( {\displaystyle\frac{1}{2}} \right)}^{{-2}}}=4\\\\ \text{(g)}\ \ {{\left( {\displaystyle\frac{{64}}{{27}}} \right)}^{{-\displaystyle\frac{2}{3}}}}={{\left( {\displaystyle\frac{{{{8}^{3}}}}{{{{3}^{3}}}}} \right)}^{{-\frac{2}{3}}}}={{\left( {\displaystyle \frac{8}{3}} \right)}^{{-2}}}={{\left( {\displaystyle \frac{3}{8}} \right)}^{2}}=\frac{9}{{64}}\\\\ \text{(h)}\ \ {{(-4)}^{{-1}}}+{{(-1)}^{{-4}}}=\displaystyle \frac{1}{{(-4)}}+\displaystyle \frac{1}{{{{{(-1)}}^{4}}}}=-\displaystyle \frac{1}{4}+1=\displaystyle \frac{3}{4} \end{array}$

2.        Simplify the following.

          (a)     $\sqrt[3]{4^{2}} \cdot 4^{\frac{2}{3}} \cdot\left(\displaystyle\frac{1}{4}\right)^{-\frac{2}{3}}$
          (b)     $\sqrt{\displaystyle\frac{512 \times 27^{-3} \times 81 \times 3^{8}}{3^{4}}}$
          (c)     $\left(\left(\displaystyle \frac{3}{4}\right)^{-4}\right)^{-0.5} \cdot \sqrt{\left(\displaystyle \frac{4}{3}\right)^{-1}} \div 16^{-0.5}$
          (d)     $(27)^{\frac{1}{4}}+\displaystyle \frac{24}{(8)^{-\frac{2}{3}}}+\frac{\sqrt[5]{2}}{(4)^{-\frac{2}{5}}}$
          (e)     $ \displaystyle \frac{(243)^{\frac{4}{5}}+(64)^{\frac{2}{3}}-(216)^{\frac{1}{3}}}{(225)^{\frac{1}{2}}-(16)^{\frac{3}{4}}}$

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$\displaystyle \begin{array}{l} \text{(a)}\ \ \ \ \sqrt[3]{{{{4}^{2}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{\left( {\frac{1}{4}} \right)}^{{-\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\cdot {{4}^{{\frac{2}{3}}}}\\ \ \ \ \ =\ {{4}^{{\frac{6}{3}}}}\\ \ \ \ \ =\ {{4}^{2}}\\ \ \ \ \ =\ 16\\\\ \text{(b)}\ \ \ \ \sqrt{{\displaystyle \frac{{512\times {{{27}}^{{-3}}}\times 81\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}\times {{3}^{{-9}}}\times {{3}^{4}}\times {{3}^{8}}}}{{{{3}^{4}}}}}}\\ \ \ \ \ =\ \sqrt{{\displaystyle \frac{{{{2}^{9}}}}{3}}}\\ \ \ \ \ =\ 16\sqrt{{\displaystyle \frac{2}{3}}}\\ \ \ \ \ =\ \displaystyle\frac{{16\sqrt{6}}}{3}\\\\ \text{(c)}\ \ \ \ {{\left( {{{{\left( {\displaystyle \frac{3}{4}} \right)}}^{{-4}}}} \right)}^{{-0.5}}}\cdot \sqrt{{{{{\left( {\frac{4}{3}} \right)}}^{{-1}}}}}\div {{16}^{{-0.5}}}\\ \ \ \ \ =\ \ {{\left( {\displaystyle \frac{3}{4}} \right)}^{2}}\cdot \sqrt{{\frac{3}{4}}}\cdot {{16}^{{0.5}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{3}^{2}}}}{{{{4}^{2}}}}\cdot \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{4}^{{\frac{1}{2}}}}}}\cdot 4\\ \ \ \ \ =\ \ \displaystyle \frac{{9\sqrt{3}}}{8}\\\\ \text{(d)}\ \ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{24}}{{{{{(8)}}^{{-\frac{2}{3}}}}}}+\frac{{\sqrt[5]{2}}}{{{{{(4)}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{{({{2}^{3}})}}^{{-\frac{2}{3}}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{{({{2}^{2}})}}^{{-\frac{2}{5}}}}}}\\ \ \ \ \ =\ \ \ {{(27)}^{{\frac{1}{4}}}}+\frac{{{{2}^{3}}\times 3}}{{{{2}^{{-2}}}}}+\frac{{{{2}^{{\frac{1}{5}}}}}}{{{{2}^{{-\frac{4}{5}}}}}}\\ \ \ \ \ =\ \ {{(27)}^{{\frac{1}{4}}}}+({{2}^{5}}\times 3)+{{2}^{{\frac{5}{5}}}}\\ \ \ \ \ =\ \ \ \sqrt[4]{{27}}+98\\\\ \text{(e)}\ \ \ \ \displaystyle \frac{{{{{(243)}}^{{\frac{4}{5}}}}+{{{(64)}}^{{\frac{2}{3}}}}-{{{(216)}}^{{\frac{1}{3}}}}}}{{{{{(225)}}^{{\frac{1}{2}}}}-{{{(16)}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{{({{3}^{5}})}}^{{\frac{4}{5}}}}+{{{({{4}^{3}})}}^{{\frac{2}{3}}}}-{{{({{6}^{3}})}}^{{\frac{1}{3}}}}}}{{{{{({{{15}}^{2}})}}^{{\frac{1}{2}}}}-{{{({{2}^{4}})}}^{{\frac{3}{4}}}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{{{3}^{4}}+{{4}^{2}}-6}}{{15-{{2}^{3}}}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{81+16-6}}{{15-8}}\\ \ \ \ \ =\ \ \ \displaystyle \frac{{91}}{7}\\ \ \ \ \ =\ \ \ 13\end{array}$

3.        Simplify the following.

         (a)     $\displaystyle \frac{x-5 \sqrt{x}}{x-2 \sqrt{x}-15} \div\left(1+\frac{3}{\sqrt{x}}\right)^{-1}$
         (b)     $\sqrt[a]{\displaystyle \frac{\sqrt[b]{x}}{\sqrt[b]{x}}} \cdot \sqrt[b]{\displaystyle \frac{\sqrt[a]{x}}{\sqrt[a]{x}}} \cdot \sqrt[c]{\frac{\sqrt[a]{x}}{\sqrt[b]{x}}}$
         (c)     $\left[\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}-y^{\frac{m}{2}}}-\displaystyle \frac{x^{m}-y^{m}}{x^{\frac{m}{2}}+y^{\frac{m}{2}}}\right]^{-2}$
         (d)     $\left(\displaystyle \frac{a^{\frac{3}{2}}}{b^{-\frac{1}{2}}}\right)^{4}\left(\displaystyle \frac{a^{-2}}{b^{3}}\right)$

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$\begin{array}{l} \text{(a)}\ \ \ \ \displaystyle \frac{{x-5\sqrt{x}}}{{x-2\sqrt{x}-15}}\div {{\left( {1+\displaystyle\frac{3}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}-5\sqrt{x}}}{{{{{\left( {\sqrt{x}} \right)}}^{2}}-2\sqrt{x}-15}}\div {{\left( {\displaystyle\frac{{\sqrt{x}+3}}{{\sqrt{x}}}} \right)}^{{-1}}}\\ \ \ \ \ =\ \ \displaystyle \frac{{\sqrt{x}\left( {\sqrt{x}-5} \right)}}{{\left( {\sqrt{x}-5} \right)\left( {\sqrt{x}+3} \right)}}\times \displaystyle \frac{{\sqrt{x}+3}}{{\sqrt{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(b)}\ \ \ \ \sqrt[a]{{\displaystyle \frac{{\sqrt[b]{x}}}{{\sqrt[c]{x}}}}}\cdot \sqrt[b]{{\displaystyle \frac{{\sqrt[c]{x}}}{{\sqrt[a]{x}}}}}\cdot \sqrt[c]{{\displaystyle\frac{{\sqrt[a]{x}}}{{\sqrt[b]{x}}}}}\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[{ab}]{x}}}{{\sqrt[{ac}]{x}}}\cdot \displaystyle\frac{{\sqrt[{bc}]{x}}}{{\sqrt[{ab}]{x}}}\cdot \displaystyle\frac{{\sqrt[{ac}]{x}}}{{\sqrt[{bc}]{x}}}\\ \ \ \ \ =\ 1\\\\ \text{(c)}\ \ \ \ {{\left[ {\displaystyle\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\frac{{{{x}^{m}}-{{y}^{m}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{{{{\left( {{{x}^{{\frac{m}{2}}}}} \right)}}^{2}}-{{{\left( {{{y}^{{\frac{m}{2}}}}} \right)}}^{2}}}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}}}-\displaystyle\frac{{\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)}}{{{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {\left( {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right)-\left( {{{x}^{{\frac{m}{2}}}}-{{y}^{{\frac{m}{2}}}}} \right)} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}-{{x}^{{\frac{m}{2}}}}+{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\ {{\left[ {2{{y}^{{\frac{m}{2}}}}} \right]}^{{-2}}}\\ \ \ \ \ =\displaystyle\frac{1}{{4{{y}^{m}}}} \end{array}$

Exponents and Radicals : Exercise (2.1) Solutions

1.       Simplify by using the rules of exponents and name the rules used.

          (a) $\displaystyle \frac{{36{{a}^{4}}{{b}^{5}}}}{{100{{a}^{7}}{{b}^{3}}}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{36{{a}^{4}}{{b}^{5}}}}{{100{{a}^{7}}{{b}^{3}}}}\\ &=\displaystyle\frac{9}{{25}}\times \frac{1}{{{{a}^{{7-4}}}}}\times {{b}^{{5-3}}}\ \ \ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{9{{b}^{2}}}}{{25{{a}^{3}}}} \end{aligned}$

          (b) $\displaystyle \frac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{27 a^{2} b^{5}}{\left(9 a^{2} b\right)^{2}}\\ &=\displaystyle \frac{{27{{a}^{2}}{{b}^{5}}}}{{81{{a}^{4}}{{b}^{2}}}}\ \ \ \ \ (\text{Power of a Powar Rule})\\ &=\displaystyle \frac{1}{3}\times \frac{1}{{{{a}^{{4-2}}}}}\times {{b}^{{5-2}}}\ \ \ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{{{b}^{3}}}}{{3{{a}^{2}}}} \end{aligned}$

          (c) $\displaystyle \left(\frac{-135 a^{4} b^{5} c^{6}}{315 a^{6} b^{7} c^{8}}\right)^{2}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle {{\left( {\frac{{-135{{a}^{4}}{{b}^{5}}{{c}^{6}}}}{{315{{a}^{6}}{{b}^{7}}{{c}^{8}}}}} \right)}^{2}}\\ &=\displaystyle {{\left( {\frac{{-3}}{{7{{a}^{{6-4}}}{{b}^{{7-2}}}{{c}^{{8-6}}}}}} \right)}^{2}}\ \ \ (\text{Division Rule})\\ &=\displaystyle \frac{{{{{(-3)}}^{2}}}}{{{{{(7)}}^{2}}{{{({{a}^{2}})}}^{2}}({{b}^{5}}){{{({{c}^{2}})}}^{2}}}}\ \ (\text{Power of a Quotient Rule})\\ &= \displaystyle \frac{9}{{49{{a}^{4}}{{b}^{5}}{{c}^{4}}}}\ \ (\text{Power of a Power Rule}) \end{aligned}$

          (d) $\displaystyle \left(\frac{x^{4}}{y^{5}}\right)^{3}\left(\frac{y^{3}}{x^{2}}\right)^{2}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle {{\left( {\frac{{{{x}^{4}}}}{{{{y}^{5}}}}} \right)}^{3}}{{\left( {\frac{{{{y}^{3}}}}{{{{x}^{2}}}}} \right)}^{2}}\\ &=\displaystyle \left( {\frac{{{{x}^{{12}}}}}{{{{y}^{{15}}}}}} \right)\left( {\frac{{{{y}^{6}}}}{{{{x}^{4}}}}} \right)\ \ (\text{Power of a Quotient Rule})\\ &=\displaystyle \frac{{{{x}^{8}}}}{{{{y}^{9}}}}\ \ (\text{Division Rule})\\ \end{aligned}$

          (e) $\displaystyle \frac{2^{3^{2}}}{\left(2^{2}\right)^{3}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{{{2}^{{{{3}^{2}}}}}}}{{{{{\left( {{{2}^{2}}} \right)}}^{3}}}}\\ &=\displaystyle \frac{{{{2}^{9}}}}{{{{2}^{6}}}}\ (\text{Power of a Power Rule})\\ &={{2}^{3}}\ \ (\text{Division Rule})\\ &=8 \end{aligned}$

2.       Evaluate the followings.

          (a) $\displaystyle \frac{54^{2} \times 12^{3} \times 64^{2}\left(3^{2} \times 4^{3} \times 5^{2}\right)^{3}}{\left(3^{2} \times 15 \times 20^{3}\right)^{4}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{{{{{54}}^{2}}\times {{{12}}^{3}}\times {{{64}}^{2}}{{{\left( {{{3}^{2}}\times {{4}^{3}}\times {{5}^{2}}} \right)}}^{3}}}}{{{{{\left( {{{3}^{2}}\times 15\times {{{20}}^{3}}} \right)}}^{4}}}}\\ &=\displaystyle \frac{{{{{\left( {{{3}^{3}}\times 2} \right)}}^{2}}\times {{{\left( {{{2}^{2}}\times 3} \right)}}^{3}}\times {{{\left( {{{2}^{6}}} \right)}}^{2}}\times {{{\left( {{{3}^{2}}\times {{2}^{6}}\times {{5}^{2}}} \right)}}^{3}}}}{{{{{\left( {{{3}^{2}}\times 3\times 5\times {{{\left( {{{2}^{2}}\times 5} \right)}}^{3}}} \right)}}^{4}}}}\\ &=\displaystyle \frac{{{{3}^{6}}\times {{2}^{2}}\times {{2}^{6}}\times {{3}^{3}}\times {{2}^{1}}^{2}\times {{3}^{6}}\times {{2}^{{18}}}\times {{5}^{6}}}}{{{{3}^{{12}}}\times {{5}^{4}}\times {{{\left( {{{2}^{2}}\times 5} \right)}}^{{12}}}}}\\ &=\displaystyle \frac{{{{3}^{{15}}}\times {{2}^{38}}\times {{5}^{6}}}}{{{{3}^{{12}}}\times {{5}^{4}}\times {{2}^{{24}}}\times {{5}^{{12}}}}}\\ &=\displaystyle \frac{{{{3}^{3}}\times {{2}^{{14}}}}}{{{{5}^{{10}}}}} \end{aligned}$

3.       Simplify.

          (a) $\displaystyle \left(\frac{3^{m}}{15^{n}}\right)^{3}\left(\frac{45^{n}}{255^{m}}\right)^{2}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \left(\frac{3^{m}}{15^{n}}\right)^{3}\left(\frac{45^{n}}{255^{m}}\right)^{2}\\ &=\displaystyle {{\left( {\frac{{{{3}^{m}}}}{{{{3}^{n}}\times {{5}^{n}}}}} \right)}^{3}}{{\left( {\frac{{{{3}^{2}}^{n}\times {{5}^{n}}}}{{{{3}^{m}}\times {{5}^{m}}\times {{{17}}^{m}}}}} \right)}^{2}}\\ &=\displaystyle \left( {\frac{{{{3}^{3}}^{m}}}{{{{3}^{3}}^{n}\times {{5}^{3}}^{n}}}} \right)\left( {\frac{{{{3}^{4}}^{n}\times {{5}^{{2n}}}}}{{{{3}^{{2m}}}\times {{5}^{{2m}}}\times {{{17}}^{{2m}}}}}} \right)\\ &=\displaystyle \frac{{{{3}^{{3m+4}}}^{n}\times {{5}^{{2n}}}}}{{{{3}^{{2m+3n}}}\times {{5}^{{2m+3n}}}\times {{{17}}^{{2m}}}}}\\ &=\displaystyle \frac{{{{3}^{{m+n}}}}}{{{{5}^{{2m+n}}}\times {{{289}}^{m}}}} \end{aligned}$

          (b) $\displaystyle \left(\frac{20^{x}}{400^{y}}\right)^{2}\left(\frac{150^{y^{2}}}{180^{x}}\right)^{3}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \left(\frac{20^{x}}{400^{y}}\right)^{2}\left(\frac{150^{y^{2}}}{180^{x}}\right)^{3}\\ &=\displaystyle {{\left( {\frac{{{{2}^{2}}^{x}\times {{5}^{x}}}}{{{{2}^{4}}^{y}\times {{5}^{{2y}}}}}} \right)}^{2}}{{\left( {\frac{{{{2}^{{{{y}^{2}}}}}\times {{3}^{{{{y}^{2}}}}}\times {{5}^{2}}^{{{{y}^{2}}}}}}{{{{2}^{2}}^{x}\times {{3}^{2}}^{x}\times {{5}^{x}}}}} \right)}^{3}}\\ &=\displaystyle {{\left( {\frac{{{{2}^{2}}^{x}\times {{5}^{x}}}}{{{{2}^{4}}^{y}\times {{5}^{{2y}}}}}} \right)}^{2}}{{\left( {\frac{{{{2}^{{{{y}^{2}}}}}\times {{3}^{{{{y}^{2}}}}}\times {{5}^{2}}^{{{{y}^{2}}}}}}{{{{2}^{2}}^{x}\times {{3}^{2}}^{x}\times {{5}^{x}}}}} \right)}^{3}}\\ &=\displaystyle \left( {\frac{{{{2}^{4}}^{x}\times {{5}^{{2x}}}}}{{{{2}^{8}}^{y}\times {{5}^{{4y}}}}}} \right)\left( {\frac{{{{2}^{{3{{y}^{2}}}}}\times {{3}^{{3{{y}^{2}}}}}\times {{5}^{6}}^{{{{y}^{2}}}}}}{{{{2}^{6}}^{x}\times {{3}^{6}}^{x}\times {{5}^{{3x}}}}}} \right)\\ &=\displaystyle {{2}^{{4x-8y}}}\cdot {{5}^{{2x-4y}}}\cdot {{2}^{{3{{y}^{2}}}}}^{{-6x}}\cdot {{3}^{{3{{y}^{2}}}}}^{{-6x}}\cdot {{5}^{{6{{y}^{2}}}}}^{{-3x}}\\ &=\displaystyle {{2}^{{3{{y}^{2}}}}}^{{-2x-8y}}\cdot {{27}^{{{{y}^{2}}}}}^{{-2x}}\cdot {{5}^{{6{{y}^{2}}}}}^{{-x-4y}} \end{aligned}$

          (c) $\displaystyle \frac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{\left(x^{3}-y^{3}\right)(x+y)}{\left(x^{2}-y^{2}\right)^{3}}\\ &=\displaystyle \frac{{(x-y)({{x}^{2}}+xy+{{y}^{2}})(x+y)}}{{{{{\left( {(x-y)(x+y)} \right)}}^{3}}}}\\ &=\frac{{(x-y)({{x}^{2}}+xy+{{y}^{2}})(x+y)}}{{{{{(x-y)}}^{3}}{{{(x+y)}}^{3}}}}\\ &=\displaystyle \frac{{{{x}^{2}}+xy+{{y}^{2}}}}{{{{{(x-y)}}^{2}}{{{(x+y)}}^{2}}}} \end{aligned}$

          (d) $\displaystyle \frac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\frac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}$

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$\begin{aligned} &\ \ \ \ \ \displaystyle \frac{\left(x^{a-b} x^{b-c}\right)^{a}\left(\frac{x^{a}}{x^{c}}\right)^{c}}{\left(x^{b} x^{c}\right)^{a} \div\left(x^{a+c}\right)^{c}}\\ &=\displaystyle \frac{{{{{\left( {{{x}^{{a-c}}}} \right)}}^{a}}{{{\left( {{{x}^{{a-c}}}} \right)}}^{c}}}}{{{{{\left( {{{x}^{{b+c}}}} \right)}}^{a}}\div {{{\left( {{{x}^{{a+c}}}} \right)}}^{c}}}}\\ &=\displaystyle \frac{{\left( {{{x}^{{{{a}^{2}}-ac}}}} \right)\left( {{{x}^{{ac-{{c}^{2}}}}}} \right)}}{{\left( {{{x}^{{ab+ac}}}} \right)\div \left( {{{x}^{{ac+c}}}^{{^{2}}}} \right)}}\\ &=\displaystyle \frac{{{{x}^{{{{a}^{2}}-{{c}^{2}}}}}}}{{{{x}^{{ab-{{c}^{2}}}}}}}\\ &=\displaystyle {{x}^{{{{a}^{2}}-ab}}} \end{aligned}$

4.       Evaluate the followings.

          (a)    $(-3)^{-2}$
          (b)    $-3^{-3}$
          (c)    $-2^{0}+5^{-1}$
          (d)    $(-2)^{-3}+2^{-2}-2^{-4}$
          (e)    $5^{0}-(-3)^{0}$
          (f)    $\displaystyle \frac{27^{-6}}{125^{-3}} \div \frac{9^{-2}}{25^{-4}}$
          (g)    $(-5)^{0}-(-5)^{-1}-(-5)^{-2}-(-5)^{-3}$
          (h)    $(-1)^{(-1)^{-1}}$
          (i)    $\displaystyle \frac{\left(180^{2}\right)^{-3}\left(6 \cdot 90^{-2}\right)^{3}}{\left(40^{-3}\right)^{2} \cdot 25^{-2}}$
          (j)    $\displaystyle \frac{\left(2^{-3}-3^{-2}\right)^{-1}}{\left(2^{-3}+3^{-2}\right)^{-1}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ {{(-3)}^{{-2}}}=\displaystyle \frac{1}{{{{{(-3)}}^{2}}}}=\displaystyle \frac{1}{9}\\\\\text{(b)}\ \ -{{3}^{{-3}}}=-\displaystyle \frac{1}{{{{3}^{3}}}}=-\displaystyle \frac{1}{{27}}\\\\\text{(c)}\ \ -{{2}^{0}}+{{5}^{{-1}}}=-1+\displaystyle \frac{1}{5}=-\displaystyle \frac{4}{5}\\\\\text{(d)}\ \ -{{2}^{0}}+{{5}^{{-1}}}=-1+\displaystyle \frac{1}{5}=-\displaystyle \frac{4}{5}\\\\\text{(e)}\ \ {{5}^{0}}-{{(-3)}^{0}}=1-1=0\\\\\text{(f)}\ \ \ \ \displaystyle \frac{{{{{27}}^{{-6}}}}}{{{{{125}}^{{-3}}}}}\div \displaystyle \frac{{{{9}^{{-2}}}}}{{{{{25}}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{({{3}^{3}})}}^{{-6}}}}}{{{{{({{5}^{3}})}}^{{-3}}}}}\div \displaystyle \frac{{{{{({{3}^{2}})}}^{{-2}}}}}{{{{{({{5}^{2}})}}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{5}^{9}}}}{{{{3}^{{18}}}}}\div \displaystyle \frac{{{{5}^{8}}}}{{{{3}^{4}}}}\\\ \ \ \ =\displaystyle \frac{{{{5}^{9}}}}{{{{3}^{{18}}}}}\times \displaystyle \frac{{{{3}^{4}}}}{{{{5}^{8}}}}\\\ \ \ \ =\displaystyle \frac{5}{{{{3}^{{14}}}}}\\\\\text{(g)}\ \ \ {{(-5)}^{0}}-{{(-5)}^{{-1}}}-{{(-5)}^{{-2}}}-{{(-5)}^{{-3}}}\\\ \ \ \ =1-\displaystyle \frac{1}{{(-5)}}-\displaystyle \frac{1}{{{{{(-5)}}^{2}}}}-\displaystyle \frac{1}{{{{{(-5)}}^{3}}}}\\\ \ \ \ =1+\displaystyle \frac{1}{5}-\displaystyle \frac{1}{{25}}+\displaystyle \frac{1}{{125}}\\\ \ \ \ =\displaystyle \frac{{125+25-5+1}}{{125}}\\\ \ \ \ =\displaystyle \frac{{146}}{{125}}\\\\\text{(h)}\ \ {{(-1)}^{{{{{(-1)}}^{{-1}}}}}}={{(-1)}^{{\displaystyle \frac{1}{{(-1)}}}}}=\ {{(-1)}^{{-1}}}=\displaystyle \frac{1}{{(-1)}}=-1\\\\\text{(i)}\ \ \ \ \displaystyle \frac{{{{{\left( {{{{180}}^{2}}} \right)}}^{{-3}}}{{{\left( {6\cdot {{{90}}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{{40}}^{{-3}}}} \right)}}^{2}}\cdot {{{25}}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{\left( {{{{\left( {{{2}^{2}}\times {{3}^{2}}\times 5} \right)}}^{2}}} \right)}}^{{-3}}}{{{\left( {3\times 2\cdot {{{(2\times {{3}^{2}}\times 5)}}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{{({{2}^{3}}\times 5)}}^{{-3}}}} \right)}}^{2}}\cdot {{{({{5}^{2}})}}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{{{{\left( {{{2}^{4}}\times {{3}^{4}}\times {{5}^{2}}} \right)}}^{{-3}}}{{{\left( {3\times 2\times {{2}^{{-2}}}\times {{3}^{{-4}}}\times {{5}^{{-2}}}} \right)}}^{3}}}}{{{{{\left( {{{2}^{{-9}}}\times {{5}^{{-3}}}} \right)}}^{2}}\cdot {{5}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-12}}}\times {{3}^{{-12}}}\times {{5}^{{-6}}}\times {{3}^{3}}\times {{2}^{3}}\times {{2}^{{-6}}}\times {{3}^{{-12}}}\times {{5}^{{-6}}}}}{{{{2}^{{-18}}}\times {{5}^{{-6}}}\times {{5}^{{-4}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-15}}}\times {{3}^{{-21}}}\times {{5}^{{-12}}}}}{{{{2}^{{-18}}}\times {{5}^{{-10}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{3}}}}{{{{3}^{{21}}}\times {{5}^{2}}}}\\\\\text{(j)}\ \ \ \ \displaystyle \frac{{{{{\left( {{{2}^{{-3}}}-{{3}^{{-2}}}} \right)}}^{{-1}}}}}{{{{{\left( {{{2}^{{-3}}}+{{3}^{{-2}}}} \right)}}^{{-1}}}}}\\\ \ \ \ =\displaystyle \frac{{{{2}^{{-3}}}+{{3}^{{-2}}}}}{{{{2}^{{-3}}}-{{3}^{{-2}}}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{{{{2}^{3}}}}+\displaystyle \frac{1}{{{{3}^{2}}}}}}{{\displaystyle \frac{1}{{{{2}^{3}}}}-\displaystyle \frac{1}{{{{3}^{2}}}}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{8}+\displaystyle \frac{1}{9}}}{{\displaystyle \frac{1}{8}-\displaystyle \frac{1}{9}}}\\\ \ \ \ =\displaystyle \frac{{\displaystyle \frac{{17}}{{72}}}}{{\displaystyle \frac{1}{{72}}}}\\\ \ \ \ =17\end{array}$

5.       Simplify the followings.

          (a)    $\left(-3 a^{4}\right)\left(4 a^{-7}\right)$
          (b)    $\left(\displaystyle \frac{2 x^{-4}}{5 y^{2} z^{3}}\right)^{-2}$
          (c)    $\left(\displaystyle \frac{x^{2 m+n} x^{3(m-n)}}{x^{m-2 n} x^{2 m-n}}\right)^{-3}$
          (d)    $\left(\displaystyle \frac{2 x^{-3} y^{2}}{3^{-1} y^{3}}\right)^{2}\left(\displaystyle \frac{4 x^{-2} y^{3}}{3 x^{5}}\right)^{3} \div\left(\frac{81 x^{-2}}{y^{-3}}\right)^{-2}$
          (e)    $ \displaystyle \frac{2 x+y}{x^{-1}+2 y^{-1}}$
          (f)    $\left(x^{-2}-y^{-1}\right)^{-3}$
          (g)    $ \displaystyle \frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$
          (h)    $\displaystyle \frac{\left(x+y^{-1}\right)^{2}}{1+x^{-1} y^{-1}}$

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$ \displaystyle \begin{array}{l}(\text{a})\ \ \left( {-3{{a}^{4}}} \right)\left( {4{{a}^{{-7}}}} \right)=-12{{a}^{{-3}}}=-\displaystyle \frac{{12}}{{{{a}^{3}}}}\\\\(\text{b})\ \ {{\left( {\displaystyle \frac{{2{{x}^{{-4}}}}}{{5{{y}^{2}}{{z}^{3}}}}} \right)}^{{-2}}}=\displaystyle \frac{{2{{x}^{{-8}}}}}{{5{{y}^{{-4}}}{{z}^{{-6}}}}}=\displaystyle \frac{{2{{y}^{4}}{{z}^{6}}}}{{5{{x}^{8}}}}\\\\(\text{c})\ \ {{\left( {\displaystyle \frac{{{{x}^{{2m+n}}}{{x}^{{3(m-n)}}}}}{{{{x}^{{m-2n}}}{{x}^{{2m-n}}}}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{{{x}^{{5m-2n}}}}}{{{{x}^{{3m-3n}}}}}} \right)}^{{-3}}}\\\ \ \ =\displaystyle \frac{{{{x}^{{-15m+6n}}}}}{{{{x}^{{-9m+9n}}}}}\\\ \ \ =\displaystyle \frac{1}{{{{x}^{{6m+3n}}}}}\\\\(\text{d})\ \ {{\left( {\displaystyle \frac{{2{{x}^{{-3}}}{{y}^{2}}}}{{{{3}^{{-1}}}{{y}^{3}}}}} \right)}^{2}}{{\left( {\displaystyle \frac{{4{{x}^{{-2}}}{{y}^{3}}}}{{3{{x}^{5}}}}} \right)}^{3}}\div {{\left( {\displaystyle \frac{{81{{x}^{{-2}}}}}{{{{y}^{{-3}}}}}} \right)}^{{-2}}}\\\ \ \ ={{\left( {\displaystyle \frac{6}{{{{x}^{3}}y}}} \right)}^{2}}{{\left( {\displaystyle \frac{{4{{y}^{3}}}}{{3{{x}^{7}}}}} \right)}^{3}}{{\left( {\displaystyle \frac{{81{{y}^{3}}}}{{{{x}^{2}}}}} \right)}^{2}}\\\ \ \ =\left( {\displaystyle \frac{{{{2}^{2}}\cdot {{3}^{2}}}}{{{{x}^{6}}{{y}^{2}}}}} \right)\left( {\displaystyle \frac{{{{2}^{6}}{{y}^{9}}}}{{{{3}^{3}}{{x}^{{21}}}}}} \right)\left( {\displaystyle \frac{{{{3}^{8}}{{y}^{6}}}}{{{{x}^{4}}}}} \right)\\\ \ \ =\displaystyle \frac{{{{2}^{8}}\cdot {{3}^{7}}\cdot {{y}^{{13}}}}}{{{{x}^{{31}}}}}\\\\(\text{e})\ \ \displaystyle \frac{{2x+y}}{{{{x}^{{-1}}}+2{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{2x+y}}{{\displaystyle \frac{1}{x}+\displaystyle \frac{2}{y}}}\\\ \ \ =\displaystyle \frac{{2x+y}}{{\displaystyle \frac{{2x+y}}{{xy}}}}\\\ \ \ =(2x+y)\left( {\displaystyle \frac{{xy}}{{2x+y}}} \right)\\\ \ \ =xy\\\\(\text{f})\ \ {{\left( {{{x}^{{-2}}}-{{y}^{{-1}}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{1}{{{{x}^{2}}}}-\displaystyle \frac{1}{y}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{y-{{x}^{2}}}}{{{{x}^{2}}y}}} \right)}^{{-3}}}\\\ \ \ ={{\left( {\displaystyle \frac{{{{x}^{2}}y}}{{y-{{x}^{2}}}}} \right)}^{3}}\\\ \ \ =\displaystyle \frac{{{{x}^{6}}{{y}^{3}}}}{{{{{\left( {y-{{x}^{2}}} \right)}}^{3}}}}\\\\\text{(g})\ \ \displaystyle \frac{{{{x}^{{-2}}}-{{y}^{{-2}}}}}{{{{x}^{{-1}}}+{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{1}{{{{x}^{2}}}}-\displaystyle \frac{1}{{{{y}^{2}}}}}}{{\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{{{{x}^{2}}{{y}^{2}}}}}}{{\displaystyle \frac{{y+x}}{{xy}}}}\\\ \ \ =\displaystyle \frac{{{{y}^{2}}-{{x}^{2}}}}{{{{x}^{2}}{{y}^{2}}}}\times \displaystyle \frac{{xy}}{{y+x}}\\\ \ \ =\displaystyle \frac{{(y-x)(y+x)}}{{{{{(xy)}}^{2}}}}\times \displaystyle \frac{{xy}}{{y+x}}\\\ \ \ =\displaystyle \frac{{y-x}}{{xy}}\\\\(\text{h})\ \ \displaystyle \frac{{{{{\left( {x+{{y}^{{-1}}}} \right)}}^{2}}}}{{1+{{x}^{{-1}}}{{y}^{{-1}}}}}\\\ \ \ =\displaystyle \frac{{{{{\left( {x+\displaystyle \frac{1}{y}} \right)}}^{2}}}}{{1+\displaystyle \frac{1}{{xy}}}}\\\ \ \ =\displaystyle \frac{{\displaystyle \frac{{{{{\left( {xy+1} \right)}}^{2}}}}{{{{y}^{2}}}}}}{{\displaystyle \frac{{xy+1}}{{xy}}}}\\\ \ \ =\displaystyle \frac{{{{{\left( {xy+1} \right)}}^{2}}}}{{{{y}^{2}}}}\times \displaystyle \frac{{xy}}{{xy+1}}\\\ \ \ =\displaystyle \frac{{x(xy+1)}}{y}\end{array}$