Asked by Anonymous
Solve the absolute value equation 4x-1=1 answers are. X=2 or x=0,, x=0or x=-1/2, x=0or x=1/2, x=2 Oryx=-1/2
Answers
Answered by
Steve
If you mean
|4x-1| = 1
then you need to recall that
|n| = n if n>=0
|n| = -n if n<0
So, we have two cases, which must be solved separately
4x-1 >= 0
4x-1=1
4x=2 x = 1/2
Since x=1/2 satisfies the condition that 4x-1 >= 0, it is a solution
4x-1 < 0
-(4x-1) = 1
-4x+1=1
-4x=0-
x=0
Since x=0 satisfies the condition that 4x-1<0, x=0 is another solution.
So, x = 0 or 1/2
This seems reasonable, since you know that |4x-1| has a v-shaped graph, which will likely intersect a horizontal line in two points.
To see this, visit
http://www.wolframalpha.com/input/?i=solve+|4x-1|%3D1
Note also that we could recognize that (x)^2 = x^2 and (-x)^2 = x^2, so we could write
(4x-1)^2 = 1^2
16x^2-8x+1=1
16x^2-8x=0
8x(2x-1) = 0
x=0 or 1/2
|4x-1| = 1
then you need to recall that
|n| = n if n>=0
|n| = -n if n<0
So, we have two cases, which must be solved separately
4x-1 >= 0
4x-1=1
4x=2 x = 1/2
Since x=1/2 satisfies the condition that 4x-1 >= 0, it is a solution
4x-1 < 0
-(4x-1) = 1
-4x+1=1
-4x=0-
x=0
Since x=0 satisfies the condition that 4x-1<0, x=0 is another solution.
So, x = 0 or 1/2
This seems reasonable, since you know that |4x-1| has a v-shaped graph, which will likely intersect a horizontal line in two points.
To see this, visit
http://www.wolframalpha.com/input/?i=solve+|4x-1|%3D1
Note also that we could recognize that (x)^2 = x^2 and (-x)^2 = x^2, so we could write
(4x-1)^2 = 1^2
16x^2-8x+1=1
16x^2-8x=0
8x(2x-1) = 0
x=0 or 1/2
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