Solve the equation `a^(2)x^(2)-3abx+2b^(2)=0` by the method of completing the square.

Explanation

We have
`a^(2)x^(2)-3abx+2b^(2)=0`
`implies" "a^(2)x^(2)-3abx=-2b^(2)`
`implies" "(ax)^(2)-2xx(ax)xx(3b)/(2)+((3b)/(2))^(2)=-2b^(2)+((3b)/(2))^(2)`
`" ""[""adding "((3b)/(2))^(2)" on both sides""]"`
`implies" "(ax-(3b)/(2))^(2)=(-2b+(9b^(2))/(4))=((-8b^(2)+9b^(2)))/(4)=(b^(2))/(4)=((b)/(2))^(2)`
`implies" "(ax-(3b)/(2))=+-(b)/(2)" "["taking square root on both sides"]`
`implies" "(ax-(3b)/(2))=(b)/(2)" or "(ax-(3b)/(2))=(-b)/(2)`
`implies" "ax=((b)/(2)+(3b)/(2))=(4b)/(2)=2b" or "ax=((-b)/(2)+(3b)/(2))=(2b)/(2)=b`
`implies" "x=(2b)/(a)" or "x=(b)/(a).`
Hence, `(2b)/(a)` and `(b)/(a)` are the roots of the given equation.