History of Hindu Mathematics: A Source Book v2.c17.s17.t4.16

v2.c17.s17.t4.16

Since & & (a\beta n - b/2)/(a\alpha) &= n, \\ we get & & n &= b/(2a(\beta - \alpha)); \\ so that & & x &= (by²)/(2a(\beta - \alpha)) + (\alpha)/a. \end{aligned} Therefore z² = a²x² + by² + c &= (b²y⁴)/(4(\beta - \alpha)²) + \alpha² + (\alpha by²)/((\beta - \alpha)) + by² + c / &= (b²y⁴)/(4(\beta - \alpha)²) + (\beta by²)/((\beta - \alpha)) + \beta² / &= { (by²)/(2(\beta - \alpha)) + \beta }². Hence z = (by²)/(2(β - α)) + β.