How do you solve # |2x-3|=x+5# and find any extraneous solutions?

1 Answer
Jan 4, 2018

Answer:

#x=8# or #x=-2/3#
(neither solution is extraneous)

Explanation:

Consider the two possibilities for #2x-3#:
#{: ("if "2x-3 >= 0,color(white)("xxx"),"if "2x-3 < 0), (abs(2x-3)=x+5,,abs(2x-3)=x+5), (rarr 2x-3=x+5,,rarr 3-2x=x+5), (rarr x=8,,rarr-2=3x), (,,rarr x=-2/3) :}#

Evaluating both results to check for extraneous solutions:
#{: ("if " x=8,,color(white)("xxxxx"),"if "x=-2/3,), (LS,=abs(2x-3),,LS,=abs(2x-3)), (,=abs(2 * (8) -3),,,=abs( 2* (-2/3)-3)), (,=abs(15),,,=abs(-13/3)), (,=13,,,=13/3), (RS,=x+5,,RS,=x+5), (,=8+5,,,=-2/3+5), (,=13,,,=13/3), (LS,=RS,,LS,=RS), ("valid",,,"valid",) :}#
Both solutions are valid (that is, neither is extraneous)