Maths Olympiad Prep

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Problem 898

AMC 12 late, AIME early
Algebra Difficulty 4.7 Prove it Autumn Tournament · Bulgaria

Solve the equation
(x+1)x2+2x+2+xx2+1=0. (x + 1)\sqrt{x^2 + 2x + 2} + x\sqrt{x^2 + 1} = 0.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Let a=x2+2x+2>0a = \sqrt{x^2 + 2x + 2} > 0 and b=x2+1>0b = \sqrt{x^2 + 1} > 0. Therefore
x=(x2+2x+2)−(x2+1)−12=a2−b2−12 x = \frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \frac{a^2 - b^2 - 1}{2}
x+1=a2−b2+12. x + 1 = \frac{a^2 - b^2 + 1}{2}.
The equation is equivalent to:
a2−b2+12⋅a+a2−b2−12⋅b=0 \frac{a^2 - b^2 + 1}{2} \cdot a + \frac{a^2 - b^2 - 1}{2} \cdot b = 0
(a2−b2)a+a+(a2−b2)b−b=0(a^2 - b^2)a + a + (a^2 - b^2)b - b = 0
(a−b)((a+b)2+1)=0.(a - b)((a + b)^2 + 1) = 0.
Hence, a=ba = b and x=−12x = -\frac{1}{2}. □\square

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