Question

abx - (a + b)(x - 1) = 0

Answer 1

Let [math]k = a/(ax-1) => x = (a+k)/k [/math][math]— (1)[/math]

Substituting for k, [math]b/(bx-1) = b / (b(a+k)/k-1) = abk/(ab+kb-ak)[/math]

Given expression is [math]k+ abk/(ab+kb-ak) = a+b[/math]

Expanding this expression, [math]abk+bk^2 - ak^2 + abk = a^2b+abk-a^2k+ab^2+kb^2-abk = a^2b-a^2k+ab^2+kb^2[/math]

=> [math]k^2(b-a)+k(2ab+a^2-b^2)-ab(a+b)=0 — (2)[/math]

Determinant for this quadratic equation is [math](2ab+a^2-b^2)^2 + 4ab(b-a)(a+b) [/math]

=[math] a^4 + b^4 + 4a^2b^2 + 4a^3b - 2a^2b^2 – 4ab^3 + 4ab^3 – 4a^3b [/math]

=[math]a^4 + b^4 + 2a^2b^2 = (a^2+b^2)^2[/math]

From equation (2), we can solve for k as [math][ -(2ab+a^2-b^2) ±(a^2+b^2) ]/2(b-a)[/math]

k will have 2 values k1 and k2 as:

[math]k1 = (-2ab-a^2+b^2 + a^2 + b^2[/math]