Question

9a²b²x² - 16abcdx - 25c²d² =0... Solve it by completing squre method​

Answer 1

$\large \sf{9 {a}^{2} {b}^{2} {x}^{2} - 16abcdx - 25 {c}^{2} {d}^{2} } = 0 \\ \\ \tt \leadsto {(3abx)}^{2} - 2(3abx) \bigg( \frac{16cd}{6} \bigg) + { \bigg( \frac{16cd}{6} \bigg)}^{2} - {\bigg( \frac{16cd}{ 6} \bigg)}^{2} - {(5cd)}^{2} = 0 \\ \\ \tt \leadsto {\bigg(3abx - \frac{16cd}{6} \bigg)}^{2} - \bigg[ \frac{16cd}{6} + 5cd\bigg] = 0\\ \\ \tt \leadsto {\bigg(3abx - \frac{16cb}{6} \bigg)}^{2} - \bigg[ \frac{16cd + 30cd}{6} \bigg] = 0 \\ \\ \tt \leadsto {\bigg(3abx - \frac{16cb}{6} \bigg)}^{2} - \frac{46cd}{6} = 0 \\ \\ \tt\leadsto \: {\bigg(3abx - \frac{16cb}{6} \bigg)}^{2} - {\bigg( \sqrt{ \frac{46cd}{6} } \bigg)}^{2} = 0 \\ \\ \tt \leadsto \bigg(3abx - \frac{16cd}{6} - \sqrt{ \frac{46cd}{6} } \bigg)\bigg(3abx - \frac{16cd}{6} + \sqrt{ \frac{46cd}{6} } \bigg ) = 0 \\ \\ \\ \bf \therefore \: \: \: \: x = \boxed{ \frac{ \frac{16cd}{6} + \sqrt{ \frac{46cd}{6} } }{3ab}} \: \: \: \: \: \: or \: \: \: \: \: \boxed{ \frac{ \frac{16cd}{6} - \sqrt{ \frac{46cd}{6} } }{3ab} }$