Grade 9 Extra exercise (Algebraic expression )

 

Exercise:

Given that,

$A\left( x \right)={{\left( 5x-1 \right)}^{2}}-{{\left( 2x+3 \right)}^{2}}$ and $B\left( x \right)=9{{x}^{2}}-24x+16-\left( 4-3x \right)\left( 3x+5 \right)$

1)    Expand and reduce $A\left( x \right)$

2)    Solve $A\left( x \right)=-8$

3)    Show that, $A\left( x \right)=\left( 7x+2 \right)\left( 3x-4 \right)$

4)    Factorize $B\left( x \right)$

5)    Given that,

$H\left( x \right)=\frac{A\left( x \right)}{B\left( x \right)}$

a)    For what values of $x$ is $H$ defined

b)    Simplify $H\left( x \right)$

c)    Solve $H\left( x \right)=2$

Solution:

1)    ${{\left( 5x-1 \right)}^{2}}={{\left( 5x \right)}^{2}}-2\left( 5x \right)\left( 1 \right)+1=25{{x}^{2}}-10x+1$

${{\left( 2x+3 \right)}^{2}}={{\left( 2x \right)}^{2}}+2\left( 2x \right)\left( 3 \right)+{{3}^{2}}=4{{x}^{2}}+12x+9$

So that, $A\left( x \right)=25{{x}^{2}}-10x+1-\left( 4{{x}^{2}}+12x+9 \right)=25{{x}^{2}}-10x+1-4{{x}^{2}}-12x-9$

                    $=\left( 25-4 \right){{x}^{2}}+\left( -10-12 \right)x+1-9=21{{x}^{2}}-22x-8$

2)    $A\left( x \right)=-8\Leftrightarrow 21{{x}^{2}}-22x=0\Leftrightarrow x\left( 21x-22 \right)=0\Leftrightarrow x=0\,\,or\,\,x=\frac{22}{21}$

3)    $A\left( x \right)=\left[ \left( 5x-1 \right)-\left( 2x+3 \right) \right]\left[ \left( 5x-1 \right)+\left( 2x+3 \right) \right]$

        $=\left( 5x-1-2x-3 \right)\left( 5x-1+2x+3 \right)=\left( 3x-4 \right)\left( 7x+2 \right)$

Or we can expand it and compare

4)    $B\left( x \right)=9{{x}^{2}}-24x+16-\left( 4-3x \right)\left( 3x+5 \right)$

Notice that, $9{{x}^{2}}-24x+16={{\left( 3x \right)}^{2}}-2\left( 3x \right)\left( 4 \right)+{{4}^{2}}={{\left( 3x-4 \right)}^{2}}$

So that, $B\left( x \right)={{\left( 3x-4 \right)}^{2}}-\left( 4-3x \right)\left( 3x+5 \right)={{\left( 3x-4 \right)}^{2}}+\left( 3x-4 \right)\left( 3x+5 \right)$

                     $=\left( 3x-4 \right)\left( 3x-4+3x+5 \right)=\left( 3x-4 \right)\left( 6x+1 \right)$

5)     a) $H$ is defined when $B\left( x \right)\ne 0\Leftrightarrow \left( 3x-4 \right)\left( 6x+1 \right)\ne 0$ i.e $x\ne \frac{4}{3}$ nor $x\ne \frac{-1}{6}$

b)$H\left( x \right)=\frac{A\left( x \right)}{B\left( x \right)}=\frac{\left( 3x-4 \right)\left( 7x+2 \right)}{\left( 3x-4 \right)\left( 6x+1 \right)}=\frac{7x+2}{6x+1}$

c) $H\left( x \right)=2\Leftrightarrow \frac{7x+2}{6x+1}=2\Leftrightarrow 7x+2=12x+2\Leftrightarrow x=0$

Grade 9 Extra Exercise (Radicals)

Exercise : Compute the following,

1)    $\sqrt{{{3}^{2}}+{{4}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}}$

2)    $\sqrt{\frac{{{8}^{10}}+{{4}^{10}}}{{{8}^{4}}+{{4}^{11}}}}$

3)    $\frac{5\sqrt{12}+\sqrt{48}}{2\sqrt{3}-\sqrt{192}-\sqrt{27}}$

4)    $\frac{{{\left( 10\sqrt{3} \right)}^{4}}}{25\times {{10}^{3}}\times 3+6\times 37.5\times {{10}^{3}}}$

5)    $\frac{0.48\times {{\left( {{10}^{3}} \right)}^{4}}\times 0.001}{0.3\times {{10}^{-4}}\times 100}$

6)    $\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{{{10}^{2}}-{{6}^{2}}}}}$

7)    $\sqrt{6-\sqrt{\frac{7}{2}+\frac{\sqrt{12}+\sqrt{27}}{\sqrt{300}}}}$

Solution:

1)    $\sqrt{{{3}^{2}}+{{4}^{2}}+{{\left( 2\sqrt{6} \right)}^{2}}}=\sqrt{9+16+{{2}^{2}}\times \sqrt{{{6}^{2}}}}=\sqrt{25+4\times 6}=\sqrt{25+24}=\sqrt{49}=\sqrt{{{7}^{2}}}=7$

2)    $\sqrt{\frac{{{8}^{10}}+{{4}^{10}}}{{{8}^{4}}+{{4}^{11}}}}=\sqrt{\frac{{{\left( 2\times 4 \right)}^{10}}+{{4}^{10}}}{{{\left( 2\times 4 \right)}^{4}}+{{4}^{11}}}}=\sqrt{\frac{{{2}^{10}}\times {{4}^{10}}+{{4}^{10}}}{{{2}^{4}}\times {{4}^{4}}+{{4}^{4+7}}}}=\sqrt{\frac{{{4}^{10}}\left( {{2}^{10}}+1 \right)}{{{4}^{4}}\left( {{2}^{4}}+{{4}^{7}} \right)}}$

                 $=\sqrt{\frac{{{4}^{10}}}{{{4}^{4}}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}+{{\left( {{2}^{2}} \right)}^{7}}}}=\sqrt{{{4}^{10-4}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}+{{2}^{14}}}}=\sqrt{{{4}^{6}}\times \frac{{{2}^{10}}+1}{{{2}^{4}}\left( 1+{{2}^{10}} \right)}}$

                 $=\sqrt{\frac{{{4}^{6}}}{{{2}^{4}}}}=\sqrt{\frac{{{\left( {{2}^{2}} \right)}^{6}}}{{{2}^{4}}}}=\sqrt{\frac{{{2}^{12}}}{{{2}^{4}}}}=\sqrt{{{2}^{12-4}}}=\sqrt{{{2}^{8}}}=\sqrt{{{\left( {{2}^{4}} \right)}^{2}}}={{2}^{4}}=16$

3)    Prime decomposition for $12={{2}^{2}}\times 3$, $48={{2}^{4}}\times 3$, $192={{2}^{6}}\times 3$ and $27=3\times {{3}^{2}}={{3}^{3}}$

$\frac{5\sqrt{12}+\sqrt{48}}{2\sqrt{3}-\sqrt{192}-\sqrt{27}}=\frac{5\sqrt{{{2}^{2}}\times 3}+\sqrt{{{2}^{4}}\times 3}}{2\sqrt{3}-\sqrt{{{2}^{6}}\times 3}-\sqrt{{{3}^{3}}}}=\frac{5\left( \sqrt{{{2}^{2}}}\times \sqrt{3} \right)+\sqrt{{{2}^{4}}}\times \sqrt{3}}{2\sqrt{3}-\sqrt{{{2}^{6}}}\times \sqrt{3}-\sqrt{{{3}^{2}}}\times \sqrt{3}}$

                           \[=\frac{\left( 5\times 2 \right)\sqrt{3}+\sqrt{{{2}^{2}}\times {{2}^{2}}}\times \sqrt{3}}{2\sqrt{3}-\sqrt{{{\left( {{2}^{3}} \right)}^{2}}}\times \sqrt{3}-3\sqrt{3}}=\frac{10\sqrt{3}+\sqrt{{{2}^{2}}}\times \sqrt{{{2}^{2}}}\times \sqrt{3}}{2\sqrt{3}-{{2}^{3}}\sqrt{3}-3\sqrt{3}}=\frac{10\sqrt{3}+4\sqrt{3}}{2\sqrt{3}-8\sqrt{3}-3\sqrt{3}}\]

\[=\frac{14\sqrt{3}}{\left( 2-8-3 \right)\sqrt{3}}=\frac{14\sqrt{3}}{-9\sqrt{3}}=-\frac{14}{9}\]

 

4)    We have, \[{{\left( 10\sqrt{3} \right)}^{4}}={{10}^{4}}\sqrt{{{3}^{4}}}={{10}^{4}}\sqrt{{{\left( {{3}^{2}} \right)}^{2}}}={{3}^{2}}\times {{10}^{4}}=9\times {{10}^{4}}\]

So that, \[\frac{{{\left( 10\sqrt{3} \right)}^{4}}}{25\times {{10}^{3}}\times 3+6\times 37.5\times {{10}^{3}}}=\frac{9\times {{10}^{4}}}{25\times {{10}^{3}}\times 3+225\times {{10}^{3}}}\]

$=\frac{9\times {{10}^{4}}}{{{10}^{3}}\left( 25\times 3+9\times 25 \right)}=\frac{9\times 10}{3\left( 25+3\times 25 \right)}=\frac{3\times 10}{25\left( 1+3 \right)}=\frac{3\times 2}{5\times 4}=\frac{3}{5\times 2}=\frac{3}{10}$

5)    $\frac{0.48\times {{\left( {{10}^{3}} \right)}^{4}}\times 0.001}{0.3\times {{10}^{-4}}\times 100}=\frac{0.48\times {{10}^{12}}\times {{10}^{-3}}}{3\times {{10}^{-1}}\times {{10}^{-4}}\times {{10}^{2}}}=\frac{48\times {{10}^{-2}}\times {{10}^{9}}}{3\times {{10}^{-5}}\times {{10}^{2}}}=\frac{48\times {{10}^{7}}}{3\times {{10}^{-3}}}=1.6\times {{10}^{11}}$

6)    $\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{{{10}^{2}}-{{6}^{2}}}}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{100-36}}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+\sqrt{64}}}$

$=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{{{2}^{3}}+8}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-\sqrt{16}}=\sqrt{{{\left( 2\sqrt{5} \right)}^{2}}-4}=\sqrt{4\times 5-4}=\sqrt{16}=4$

7)     We have $\sqrt{12}=\sqrt{{{2}^{2}}\times 3}=\sqrt{{{2}^{2}}}\times \sqrt{3}=2\sqrt{3}$,    

      $\sqrt{27}=\sqrt{{{3}^{2}}\times 3}=\sqrt{{{3}^{2}}}\times \sqrt{3}=3\sqrt{3}$ and $\sqrt{300}=\sqrt{3\times 100}=\sqrt{3}\times \sqrt{100}=10\sqrt{3}$

So that, \[\sqrt{6-\sqrt{\frac{7}{2}+\frac{\sqrt{12}+\sqrt{27}}{\sqrt{300}}}}=\sqrt{6-\sqrt{\frac{7}{2}+\frac{5\sqrt{3}}{10\sqrt{3}}}}=\sqrt{6-\sqrt{\frac{7}{2}+\frac{1}{2}}}=\sqrt{6-\sqrt{4}}\]

                                              \[=\sqrt{6-2}=\sqrt{4}=2\]