How do you solve and check for extraneous solutions in #abs(x-1) = 5x + 10#?

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Answer 1 · 2015-08-02T19:42:33

Solution: #x = -3/2#
Extraneous solution: #x = -11/4#

If you take into account the fact that the absolute value of a number is always positive regardless if said number is positive or negative

#color(blue)( |n| = {(n",", "if "n>=0), (-n",", "if "n<0) :})#

then you can say that the solutions to this equation must satisfy the condition

#5x+10 >0 <=> x> -2#

Now, your absolute value equation will produce two solutions, depending on which condition is true

#|x-1| = x-1#

This will get you

#x-1 = 5x+10 => x = color(red)(-11/4)#

#|x-1| = -(x-1) = -x+1#

The solution to the equation will be

#-x+1 = 5x + 10 => x = -9/6 = color(green)(-3/2)#

As you can see, #x = -11/4# does not satisfy the condition #x> -2#, which means that this solution will be extraneous.

The only solution to this equation will thus be #x = -3/2#.