Find the roots of the equation $a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$ by the method of competing the square.

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Solution

We have
$a^{2} x^{2} - 3 a b x + 2 b^{2} = 0$

$\Rightarrow x^{2} - \frac{3 b}{a} x + 2 \frac{b^{2}}{a^{2}} = 0 \Rightarrow x^{2} - \frac{3 b}{a} x = - 2 \frac{b^{2}}{a^{2}}$

$\Rightarrow x^{2} - 2 (\frac{3 b}{2 a}) x + {(\frac{3 b}{2 a})}^{2} = - \frac{{2 b}^{2}}{a^{2}} + {(\frac{3 b}{2 a})}^{2} \Rightarrow {(x - \frac{3 b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}}$

$\Rightarrow x = \frac{3 b}{2 a} + \frac{b}{2 a} = \frac{2 b}{a} o r x = \frac{3 b}{2 a} - \frac{b}{2 a} = \frac{b}{a}$

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