solve for x: a/ax-1 + b/bx-1 = a+b, where x is not = 1/a, 1/b a+b is not = - Maths - Quadratic Equations

Question

solve for x: a/ax-1 + b/bx-1 = a+b, where x is not = 1/a, 1/b
a+b is not = 0, ab is not = 0

Harleen Birdi · Accepted Answer

$\frac{a}{a x - 1} + \frac{b}{b x - 1} = a + b \Rightarrow a (b x - 1) + b (a x - 1) = (a + b) (a x - 1) (b x - 1) \Rightarrow a b x - a + a b x - b = a^{2} b x^{2} - a^{2} x - a b x + a + a b^{2} x^{2} - a b x - b^{2} x + b \Rightarrow (a^{2} b + a b^{2}) x^{2} - (a^{2} + b^{2} + 4 a b) x + 2 (a + b) = 0 D = (a^{2} + b^{2} + 4 a b)^{2} - 4 \times (a^{2} b + a b^{2}) \times 2 (a + b) = a^{4} + b^{4} + 16 a^{2} b^{2} + 2 a^{2} b^{2} + 8 a b^{3} + 8 a^{3} b - 8 a^{3} b - 8 a^{2} b^{2} - 8 a^{2} b^{2} - 8 a b^{3} = a^{4} + b^{4} + 2 a^{2} b^{2} = (a^{2} + b^{2})^{2} \Rightarrow \sqrt{D} = (a^{2} + b^{2})$

$∴ x = \frac{(a^{2} + b^{2} + 4 a b) \pm \sqrt{D}}{2 (a^{2} b + a b^{2})} = \frac{a^{2} + b^{2} + 4 a b \pm (a^{2} + b^{2})}{2 a b (a + b)} \Rightarrow x = \frac{a^{2} + b^{2} + 4 a b + a^{2} + b^{2}}{2 a b (a + b)}, \frac{a^{2} + b^{2} + 4 a b - a^{2} - b^{2}}{2 a b (a + b)} = \frac{2 (a^{2} + b^{2} + 2 a b)}{2 a b (a + b)}, \frac{4 a b}{2 a b (a + b)} = \frac{(a + b)^{2}}{a b (a + b)}, \frac{2}{a + b} = \frac{a + b}{a b}, \frac{2}{a + b}$

Hence, two roots of x are $\frac{a + b}{a b}$ and $\frac{2}{a + b}$

solve for x: a/ax-1 + b/bx-1 = a+b, where x is not = 1/a, 1/b. a+b is not = 0, ab is not = 0