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