Question

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

Answer 1

`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`