Question

Find the roots of the quadratic equation a²x²-3abx+2b²=0 but the method of completing the square

Answer 1

this is your answer. check it

Answer 2

Given : $\sf{ {a}^{2} {x}^{2} - 3abx + {2b}^{2} = 0}$

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

➡️$\sf{ {x}^{2} - \frac{3b}{a} x = - \frac{ {2b}^{2} }{ {a}^{2} } }$

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

➡️$\sf{ {(x - \frac{3b}{2a}) }^{2} = \frac{ {9b}^{2} }{ {4a}^{2} } - \frac{ {2b}^{2} }{ {a}^{2} } = \frac{ {9b}^{2} - {8b}^{2} }{ {4a}^{2} } = \frac{ {b}^{2} }{ {4a}^{2} } }$

➡️$\sf{x - \frac{3b}{2a} = + \frac{b}{2a} = x = \frac{3b}{2a} + \frac{b}{2a}}$

➡️$\sf{x = \frac{3b + b}{2a} \: or \: x = \frac{3b - b}{2a}}$

➡️$\sf{x = \frac{2b}{a} \: or \: x = \frac{b}{a}}$

Hence, the roots of the given equation are $\sf{\frac{2b}{a}and\frac{b}{a}}$