How do you solve #abs(2t-3) = t# and find any extraneous solutions?

1 Answer
May 4, 2018

Answer:

#t=1# or #t=3# and despite squaring equations, no extraneous solutions suggested themselves.

Explanation:

Squaring usually introduces extraneous solutions. It's worth it because it turns the whole thing to straightforward algebra, eliminating the confusing case analysis typically associated with an absolute value question.

# (2t-3)^2 = t^2#

#4t^2 - 12 t + 9 = t^2#

#3(t^2 -4t + 3) = 0#

# (t-3)(t-1)=0#

#t=3# or #t=1#

We're in good shape because no negative #t# values came up, which are surely extraneous, We'll check these two but they should be OK.

# |2(3) - 3| = |3|=3=t quad sqrt#

#|2(1)-3| = |-1|=1=t quad sqrt#