Question 6
The number of real solutions of the equation $$\frac{(x+2)(x+3)(x+4)(x+5)}{(x-2)(x-3)(x-4)(x-5)} = 1$$ is
Solution
Pairing the factors, the left side numerator is $$(u + 10)(u + 12)$$ with $$u = x^2 + 7x$$ and the denominator is $$(v + 10)(v + 12)$$ with $$v = x^2 - 7x$$. The equation becomes $$u^2 + 22u = v^2 + 22v$$, that is $$(u - v)(u + v + 22) = 0$$. Since $$u - v = 14x$$ and $$u + v + 22 = 2x^2 + 22 > 0$$, the only real solution is $$x = 0$$.
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