History of Hindu Mathematics: A Source Book v2.c17.s22.44

v2.c17.s22.44

The auxiliary equation abr² + a²d - abc = s² in this case becomes abr² + (a²β² - aba²) = s². The same equation is obtained by proceeding as in case (*i*) with the assumption v = by + β. An obvious solution of it is r = α, s = aβ. Hence in this case the general solution (1.3) becomes x &= 1/(a(t² - ab)²){\alpha(t² + ab) ± 2a\beta t}² - (\alpha²)/c, / u &= 1/((t² - ab)){\alpha(t² + ab) ± 2a\beta t}, / v &= 1/((t² - ab)){\beta(t² + ab) ± 2bat}, where t is any rational number.