Mathematics in Ancient and Medieval India c6.s5.20

c6.s5.20

Squaring and neglecting e₁², we get $2((2a²+a+r)/(2a+1)).₁=(a²+r)-((2a²+a+r)/(2a+1))²$ $=(r(2a+1-r))/((2a+1)²)$ $₁=(r(2a+1-r))/(2(2a+1)(2a²+a+r))=(r(2a+1-r))/(2(2a+1)²(a+r/(2a+1)))$ $=(r/(2a+1)(1-r/(2a+1)))/(2(a+r/(2a+1)))$