Question

[b - (b - x^2)^1/2] / [b + (b - x^2)^1/2] = a

Answer 1

Is the question written properly ??   is that b or  b^2  under the root...?

Let     (a+b+1-ab)/2 = k       a constant.
         (1+a)/2b = m    a constant

$\frac{b-\sqrt{b-x^2}}{b+\sqrt{b-x^2}}=a\\\\Rationalize\ Nr\ and\ Dr\\\\\frac{b-\sqrt{b-x^2}}{b+\sqrt{b-x^2}}*\frac{b-\sqrt{b-x^2}}{b-\sqrt{b-x^2}}=a\\\\\frac{b^2+b-x^2-2b\sqrt{b-x^2}}{b^2-(b-x^2)}=a\\\\2b\sqrt{b-x^2}=b^2+b-x^2-ab^2+ab-ax^2\\\\\sqrt{b-x^2}=k -m x^2\\\\b-x^2=k^2+m^2x^4-2kmx^2\\\\m^2 x^4-(2km-1)x^2+k^2-b=0\\\\x^2=\frac{2km-1+-\sqrt{(2km-1)^2-4m^2(k^2-b)}}{2m^2}\\\\x^2=\frac{2km-1+-\sqrt{1-4km+4m^2b}}{2m^2}$