An algebra problem by Aly Ahmed

Algebra Level pending

A A and B B are the solutions to ⌊ x 2 2 ⌋ = ⌊ 2 x 2 ⌋ \left \lfloor \dfrac {x^2}2 \right \rfloor = \left \lfloor \dfrac 2{x^2} \right \rfloor

Submit your answer as A + B A+B .


The answer is 0.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

Tom Engelsman
Jun 14, 2020

The quick & dirty solution occurs when the two arguments are equal. This leads to:

x 2 2 = 2 x 2 ⇒ x 4 = 4 ⇒ ( x 2 − 2 ) ( x 2 + 2 ) = 0 ⇒ x = ± 2 \frac{x^2}{2} = \frac{2}{x^2} \Rightarrow x^4 = 4 \Rightarrow (x^2-2)(x^2+2) = 0 \Rightarrow x = \pm \sqrt{2}

are the only real solutions for A A and B B . Hence, A + B = 0 . A+B=\boxed{0}.

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