Question

Solve by completing the square method
2q - 3 / 5 = 2 (2q-3) / 3q.

Answer 1

$2q - \frac{3}{5} = \frac{2(2q - 3)}{3q} \\ \frac{10q - 3}{5} = \frac{4q - 6}{3q} \\ (10q - 3)(3q) = (4q - 6)(5) \\ 30 {q}^{2} - 9q = 20q - 30 \\ 30 {q}^{2} - 9q - 20q + 30 = 0 \\ 30 {q}^{2} - 29q + 30 = 0 \\ \\ solving \: by \: completing \: square \: \\ we \: take \: the \: constant \: to \: rhs \: and \: make \: \\ coefficient \: of \: {q}^{2} as \: 1 \\ \\ 30 {q}^{2} - 29q = - 30 \\ \\ dividing \: throughout \: by \: 30 \\ \\ {q}^{2} - \frac{29}{30} q = - 1 \\ \\ adding \: {( \frac{1}{2} b)}^{2} = \frac{ {b}^{2} }{4} = \frac{ { (\frac{ - 29}{30} )}^{2} }{4} = \frac{ \frac{841}{900} }{4} = \frac{841}{3600} \\ on \: both \: sides \\ \\ {q}^{2} - \frac{29}{30} q + \frac{841}{3600} = - 1 \\ {(q)}^{2} - 2(q)( \frac{29}{60} ) + {( \frac{29}{60}) }^{2} = - 1 \\ {(q - \frac{29}{60}) }^{2} = - 1 \\ q - \frac{29}{60} = + \sqrt{ - 1} \: \: \: \: or \: \: \: - \sqrt{ - 1} \\ q = \sqrt{ - 1} + \frac{29}{60} \: \: \: \: or \: \: \: \: \frac{29}{60} - \sqrt{ - 1} \\ \\ \sqrt{ - 1} \: is \: an \: imaginary \: number \\ hence \: the \: roots \: of \: the \:given \: equations \: are \: \\ complex$

*if you're in 11th standard or higher, the further steps apply for you to solve too*

$q = \frac{29}{60} + \sqrt{ - 1} \: \: \: \: or \: \: \: \: \frac{29}{60} - \sqrt{ - 1} \\ q = \frac{29}{60} + i \: \: \: \: \: or \: \: \: \: \: \frac{29}{60} - i\: \: \: \: \: \: \: \: \: \: \: \: ... \: (i = \sqrt{ - 1} ) \\ q = \frac{29 + 60i}{60} \: \: \: \: \: or \: \: \: \: \: \frac{29 - 60i}{60}$