[5 marks sum]

Question 15 Marks

Solve the following quadratic equation using formula method only
$3a^2x^2 +8abx + 4b^2 = 0, a \neq 0$

Answer

$3 a^2 x^2+8 a b x+4 b^2=0$
$ x ^2+\frac{8 b }{3 a } x +\frac{4 b ^2}{3 a ^2}=0$
$ a=1 ; b=\frac{8 b}{3 a} ; c=\frac{4 b^2}{3 a^2}$
$ D=b^2-4 a c \\ =\left(\frac{8 b }{3 a }\right)^2-4(1)\left(\frac{4 b ^2}{3 a ^2}\right) $
$=\frac{64 b^2}{9 a^2}-\frac{16 b^2}{3 a^2} $
$=\frac{64 b^2-48 b^2}{9 a^2}=\frac{16 b^2}{9 a^2} $
$ x =\frac{- b \pm \sqrt{ b ^2-4 ac }}{2 a} $
$ x=\frac{-\frac{8 b }{3 a } \pm \sqrt{\frac{16 b ^2}{9 a ^2}}}{2} $
$ x=\frac{-\frac{8 b}{3 a}+\frac{4 b}{3 a}}{2}, x=\frac{-\frac{8 b}{3 a}-\frac{4 b}{3 a}}{2}$
$x=\frac{-4 b}{6 a}, x=\frac{-12 b}{6 a}$
$x=-\frac{2 b}{3 a}, x=-\frac{2 b}{a} $

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Question 25 Marks

Solve the following equation :
$\frac{ x -1}{ x -2}+\frac{ x -3}{ x -4}=3 \frac{1}{3}$

Answer

$\frac{ x -1}{ x -2}+\frac{ x -3}{ x -4}=3 \frac{1}{3} $
$ \frac{( x -1)( x -4)+( x -3)( x -2)}{( x -2)( x -4)}=\frac{10}{3} $
$ \frac{x^2-5 x+4+x^2-5 x+6}{x^2-6 x+8}=\frac{10}{3}$
$ \frac{2 x^2-10 x+10}{x^2-6 x+8}=\frac{10}{3} $
$ 6 x^2-30 x+30=10 x^2-60 x+80$
$4 x^2-30 x+50=0$
$ 2 x^2-15 x+25=0 $
$x ^2-\frac{15}{2} x +\frac{25}{2}=0$
$x^2-5 x-\frac{5}{2} x+\frac{25}{2}=0 $
$x(x-5)-\frac{5}{2}(x-5)=0$
$ (x-5)\left(x-\frac{5}{2}\right)=0 $
$ x=5, x=\frac{5}{2}$

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Question 35 Marks

Solve the following equation: $\frac{ a }{ x - a }+\frac{ b }{ x - b }=\frac{2 c }{ x - c }$

Answer

$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2 c}{x-c}$
$\frac{a(x-b)+b(x-a)}{(x-a)(x-b)}=\frac{2 c}{x-c}$
$\frac{a x-a b+b x-a b}{x^2-a x-b x+a b}=\frac{2 c}{x-c}$
$ \frac{(a+b) x-2 a b}{x^2-(a+b) x+a b}=\frac{2 c}{x-c}$
$ \{(a+b) x-2 a b\}(x-c)=2 c\left\{x^2-(a+b) x+a b\right\} $
$ (a+b) x^2-2 a b x-c(a+b) x+2 a b c=2 c x^2-2 c(a+b) x+2 a b c $
$ (a+b) x^2-[2 a b+c(a+b)] x+2 a b c=2 c x^2-2 c(a+b) x+2 a b c $
$ (a+b-2 c) x^2=(2 a b+a c+b c-2 c a-2 b c) x$
$ (a+b-2 c) x^2=(2 a b-a c-b c) x $
$ x=0, x=\frac{2 a b-a c-b c}{a+b-2 c}$

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Question 45 Marks

Solve the following equation:
$\frac{1}{( x -1)(x-2)}+\frac{1}{( x -2)( x -3)}+\frac{1}{( x -3)( x -4)}=\frac{1}{6}$

Answer

$\frac{1}{( x -1)(x-2)}+\frac{1}{( x -2)( x -3)}+\frac{1}{( x -3)( x -4)}=\frac{1}{6}$
$\frac{( x -3)( x -4)+( x -1)( x -4)+( x -1)( x -2)}{( x -1)( x -2)( x -3)( x -4)}=\frac{1}{6} $
$\frac{ x ^2-3 x -4 x +12+ x ^2- x -4 x +4+ x ^2- x -2 x +2}{( x -1)( x -2)( x -3)( x -4)}=\frac{1}{6} $
$ \frac{3 x ^2-15 x +18}{( x -1)( x -2)( x -3)( x -4)}=\frac{1}{6} $
$ \frac{3\left( x ^2-5 x +6\right)}{( x -1)( x -2)( x -3)( x -4)}=\frac{1}{6} $
$ \frac{3( x -3)( x -2)}{( x -1)( x -2)( x -3)( x -4)}=\frac{1}{6}$
$\frac{3}{(x-1)(x-4)}=\frac{1}{6} $
$x^2-5 x+4=18 $
$ x^2-5 x-14=0 $
$x^2+2 x-7 x-14=0 $
$ x(x+2)-7(x+2)=0 $
$ (x+2)(x-7)=0$
$ x=-2, x=7$

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Mathematics [5 marks sum]

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