Asked by george
I am asked to solve the absolute value equation of
|1/2x +1| =x+1 graphically. I can solve this algebraically and consider two cases for the absolute value one being positive and the other negative. I know this will be a straight line and when I solved it algebraically x=0 the other solution is -4/3 which is an extraneous root and does not work. How do I graph this? Thanks for your help
|1/2x +1| =x+1 graphically. I can solve this algebraically and consider two cases for the absolute value one being positive and the other negative. I know this will be a straight line and when I solved it algebraically x=0 the other solution is -4/3 which is an extraneous root and does not work. How do I graph this? Thanks for your help
Answers
Answered by
Reiny
from the definition of absolute value, we know the right side must be ≥0
so x+1 ≥ 0
x ≥ -1
algebraically:
(1/2)x + 1 = x+1 OR (-1/2)x - 1 = x+1
x + 2 = 2x + 2 OR -x - 2 = 2x + 2
x = 0 OR -3x = 4
x = 0 or x = -4/3
which is what you had
remember at the top we said x ≥ -1 , so yes, the -4/3 has to be rejected.
to see this graphically
|1/2x +1| -x-1 = 0
let y = |1/2x +1| -x-1
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7C1%2F2x+%2B1%7C+-x-1
see where it crosses the x-axis ??
only at x = 0
so x+1 ≥ 0
x ≥ -1
algebraically:
(1/2)x + 1 = x+1 OR (-1/2)x - 1 = x+1
x + 2 = 2x + 2 OR -x - 2 = 2x + 2
x = 0 OR -3x = 4
x = 0 or x = -4/3
which is what you had
remember at the top we said x ≥ -1 , so yes, the -4/3 has to be rejected.
to see this graphically
|1/2x +1| -x-1 = 0
let y = |1/2x +1| -x-1
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7C1%2F2x+%2B1%7C+-x-1
see where it crosses the x-axis ??
only at x = 0
Answered by
Steve
perhaps it will help to view the two functions graphed individually:
http://www.wolframalpha.com/input/?i=+|1%2F2x+%2B1|+%3D+x%2B1
http://www.wolframalpha.com/input/?i=+|1%2F2x+%2B1|+%3D+x%2B1
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