Source: 2010 Saudi Arabia Pre-TST 3.4
Tags: parameterization, algebra
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Let $a$ and $b$ be real numbers such that $a + b \ne 0$. Solve the equation
$$\frac{1}{(x + a)^2 - b^2} +\frac{1}{(x +b)^2 - a^2}=\frac{1}{x^2 -(a + b)^2}+\frac{1}{x^2-(a -b)^2}$$
$\frac{1}{(x+a)^2 - b^2} + \frac{1}{(x+b)^2 - a^2} = \frac{1}{x^2 - (a+b)^2} + \frac{1}{x^2 - (a-b)^2}$;
or $\frac{1}{(x+a+b)(x+a-b)} + \frac{1}{(x+a+b)(x-a+b)} = \frac{1}{(x-a-b)(x+a+b)} + \frac{1}{(x-a+b)(x+a-b)}$ (i).
Let us next multiply both sides of (i) by $(x + a + b)(x + a - b)(x - a + b)(x - a - b)$ to obtain:
$(x-a+b)(x-a-b) + (x+a-b)(x-a-b) = (x+a-b)(x-a+b) + (x+a+b)(x-a-b)$;
or $(x - a)^2 - b^2 + (x - b)^2 - a^2 = x^2 - (a - b)^2 + x^2 - (a + b)^2$;
or $x^2 - 2ax + a^2 - b^2 + x^2 - 2bx + b^2 - a^2 = x^2 - (a^2 - 2ab + b^2) + x^2 - (a^2 + 2ab + b^2)$;
or $-2(a + b)x + 2x^2 = 2x^2 - 2(a^2 + b^2)$;
or $-2(a + b)x = -2(a^2 + b^2)$;
or $x = \frac{a^2 + b^2}{a + b}$.