History of Hindu Mathematics: A Source Book v2.c17.s17.t1.13

v2.c17.s17.t1.13

Completing the square on the left-hand side of the equation ax² + bx + c = y², we have $(ax + ½b)² = ay² + ¼(b² - 4ac).$ Putting z = ax + ½b, k = ¼(b² - 4ac), we get $ay² + k = z². 1.1$ If y = l, z = m be found empirically to be a solution of this equation, another solution of it will be $y = lq ± mp,$ $z = mq ± alp,$ where ap² + 1 = q². Hence a solution of (1) is $x = - b/2a + 1/a(mq ± alp),$ $y = lq ± mp.$