Question

a2x2-3abx+2b2=0
$a2x2 - 3abx + 2b = 0$
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Answer 1

Answer: Hi ,

a² x² -3abx + 2b² = 0

a² x² - abx - 2abx + 2b² = 0

ax( ax - b ) - 2b( ax - b ) = 0

( ax - b )( ax -2b ) = 0

Therefore ,

( ax -b ) = 0 or ax -2b = 0

ax = b or ax = 2b

x = b / a or x = 2b / a

I hope this helps you.

:)

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 $\sf{\frac{2b}{a}and\frac{b}{a}}$