How do you solve #-x = sqrt(2x+15)# and find any extraneous solutions?

1 Answer
Y
May 7, 2017

Solve for x then substitute the values in to verify that they are not extraneous solutions; x=5 is extraneous.

Explanation:

First, we solve for x:
#(-x)^2=2x+15#
#x^2-2x-15=0#
#(x-5)(x+3)=0#
#x=-3, 5#

If #x=-3#
#-(-3)=sqrt(2(-3)+15)#
#3=sqrt(-6+15)=sqrt(9)=3#
therefore #x=-3# is not extraneous

If #x=5#
#-5=sqrt(2(5)+15)#
Since the square root of a number cannot be negative, #x=5# is an extraneous solution.

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