( a + 2 ) 2 + ( a 2 ) 2 = 4 \sqrt {(a+2)^2} + \sqrt {(a-2)^2} = 4

Algebra Level 3

( a + 2 ) 2 + ( a 2 ) 2 = 4 \displaystyle \sqrt {(a+2)^2} + \sqrt {(a-2)^2} = 4 , solve for the range of a a .

1 a 1 -1 \leq a \leq 1 0 2 a 2 -2 \leq a \leq 2 1 < a < 1 -1 < a < 1 2 < a < 2 -2 < a < 2

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2 solutions

Lingga Musroji
Jan 5, 2021

( a + 2 ) 2 + ( a 2 ) 2 = 4 a + 2 + a 2 = 4 \sqrt{(a+2)^2}+\sqrt{(a-2)^2}=4\\|a+2|+|a-2|=4

Case 1 2 a 2 -2\leq a\leq 2 :

( a + 2 ) ( a 2 ) = 4 4 = 4 (a+2)-(a-2)=4\\4=4 is always true

Case 2 a > 2 a>2 : ( a + 2 ) + ( a 2 ) = 4 2 a = 4 a = 2 (a+2)+(a-2)=4\\2a=4\\a=2 no solution since a > 2 a>2

Case a < 2 a<-2 : ( a + 2 ) ( a 2 ) = 4 2 a = 4 a = 2 -(a+2)-(a-2)=4\\-2a=4\\a=-2 no solution since a < 2 a<-2

Hence, the solution is 2 a 2 -2\leq a\leq 2

Wow, a very good solution.

. . - 5 months, 1 week ago
. .
Jan 4, 2021

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