Question

find the roots of the above equation by the method of completing square PLEASE ITS URGENT
I will be comfortable if the answer is in the form of an attachment
I will MARK YOU AS THE BRAINLIEST!
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Answer 1

Step-by-step explanation:

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Answer 2

SOLUTION

$= > {a}^{2} {x}^{2} - 3abx = - 2 {b}^{2} \\ = > ( {ax)}^{2} - 2 \times ax \times \frac{3b}{2} = - 2 {b}^{2} \\ \\ = > ( {ax)}^{2} - 2 \times ax \times \frac{3b}{2} + ( \frac{3b}{2} ) {}^{2} = - 2 {b}^{2} + ( \frac{3b}{2} ) {}^{2} \\ \\ = > (ax - \frac{3b}{2} ) {}^{2} = - 2 {b}^{2} + \frac{9 {b}^{2} }{4} \\ \\ = > (ax - \frac{3b}{2} ) {}^{2} = \frac{ {b}^{2} }{4} \\ \\ = > ax - \frac{3b}{2} = </u><u>±</u><u> \frac{b}{2} \\ \\ = > ax = \frac{3b}{2} </u><u>±</u><u> \frac{b}{2} \\ \\ = > ax =</u><u> \frac{3b}{2} + \frac{b}{2} \: \: \: or \: \: \: ax = \frac{3b}{2} - \frac{b}{2} \\ \\ = > ax = 2b \: \: \: \: or \: \: \: \: ax = b \\ \\ = > x = \frac{2b}{a} \: or \: \: \: x = \frac{b}{a}$

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