How do you solve and check for extraneous solutions in #abs(2x + 3) = 5#?

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Answer 1 · 2015-08-01T12:53:56

Solve: |2x + 3| = 5

Ans : x = 1, x = - 4

separate solving in 2 cases:
a. (2x + 3) = 5 --> 2x = 2 --> #x = 1#

b. - (2x + 3 ) = 5 --> -2x - 3 = 5 --> -2x = 8 --># x = - 4#

Answer 2 · 2015-08-01T14:57:29

#color(red)(x = 1)# and # color(red)(x = -4)#.
There are #color(red)(" no")# extraneous solutions.

SOLVE

We need to write two different equations without the absolute value symbols and solve for #x#.

These equations are
(1): #(2x+3) = 5#
(2): #-(2x+3) = 5#

Solve Equation 1:

#2x+3 = 5#

Subtract #3# from each side.

#2x = 2#

Divide each side by #2#.

#x = 1#

Solve Equation 2:

#−(2x+3) = 5#

Remove parentheses.

#−2x−3= 5#

Add #3# to each side.

#-2x = 8#

Divide each side by #-2#.

#x = -4#

The solutions are #x = 0# and #x = -4#.

CHECK FOR EXTRANEOUS SOLUTIONS:

If #x = 1#,

#|2x+3|=5#
#|2×1+3|= 5#
#|2+3| = 5#
#|5| =5#
#5=5#

#x=1# is a true solution.

If #x= -4#,

#|2x+3|=5#
#|2(-4) + 3| = 5#
#|-8 + 3| = 5#
#|-5| = 5#
#5 = 5#

#x=5# is a true solution.

There are no extraneous solutions.