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

May 4, 2018

$t = 1$ 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.

${\left(2 t - 3\right)}^{2} = {t}^{2}$

$4 {t}^{2} - 12 t + 9 = {t}^{2}$

$3 \left({t}^{2} - 4 t + 3\right) = 0$

$\left(t - 3\right) \left(t - 1\right) = 0$

$t = 3$ 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