An algebra problem by gaoyi zhu

Algebra Level 2

Given that a < 1 a<-1 , which of the following is equal to a 2 + ( a + 1 ) 2 ? \sqrt{a^2} + \sqrt{(a+1)^2} ?

1 1 1 + 2 a 1+2a 1 -1 1 2 a -1-2a

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1 solution

Gaoyi Zhu
Mar 25, 2020

Since the square root of a ^2 in the expression is a positive square root, then sqrt( a ^2) > 0. If a < 0, then - a > 0, which implies that sqrt( a ^2)= - a . Similarly, the square root of ( 1- a )^2 is a positive square root too, so sqrt[( 1- a )^2] > 0. If a <0, then 1- a >0, which implies that sqrt[(1- a )^2]=1- a . Therefore the expression above could be simplified into - a + 1- a = 1-2a .

Did you all get tricked?

gaoyi zhu - 1 year, 2 months ago

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Uhm, excuse me? There is no 1 2 a 1-2a in your answer?

Steven Jim - 1 year, 2 months ago

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