Long Equation

Algebra Level pending

Find all real values of x that satisfies this equation.


The answer is -1.

William Isoroku by William Isoroku , 25, USA

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1 solution

Tom Engelsman
Aug 7, 2020

Let u = 1 − x 2 − 1 − x 2 + 2 x + 1 u = 1 - \sqrt{x^2-1} - \sqrt{x^2+2x+1} such that u + 1 u = 2 ⇒ u 2 − 2 u + 1 = ( u − 1 ) 2 = 0 ⇒ u = 1. u + \frac{1}{u} = 2 \Rightarrow u^2 - 2u + 1 = (u-1)^2 = 0 \Rightarrow u = 1. Thus, we now have:

1 − x 2 − 1 − x 2 + 2 x + 1 = 1 ⇒ − x 2 − 1 = x 2 + 2 x + 1 ⇒ x 2 − 1 = x 2 + 2 x + 1 ⇒ 2 x = − 2 ⇒ x = − 1 . 1 - \sqrt{x^2-1} - \sqrt{x^2+2x+1} = 1 \Rightarrow - \sqrt{x^2-1} = \sqrt{x^2+2x+1} \Rightarrow x^2 - 1 = x^2 + 2x + 1 \Rightarrow 2x = -2 \Rightarrow \boxed{x=-1}.

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