4The roots are 3 and 1.Trial and error is practical and quicker, but the mathematical solution:|x-2|² + |x-2| - 2 = 0a²+a-2 = 0, a = |x-2|a = 1, -2Since a is an absolute value, -2 is not a solution. Thus, a = 1|x-2| = 1x-2 = 1, -1x = 3, 1
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ReplyDeleteThe roots are 3 and 1.
Trial and error is practical and quicker, but the mathematical solution:
|x-2|² + |x-2| - 2 = 0
a²+a-2 = 0, a = |x-2|
a = 1, -2
Since a is an absolute value, -2 is not a solution. Thus, a = 1
|x-2| = 1
x-2 = 1, -1
x = 3, 1