How do you solve and check for extraneous solutions in #x^2 + 2 = x + 4#?

1 Answer
Aug 7, 2015

Answer:

Express the given equation in standard form, factor for solution values, and check each resulting value to ensure that it is not extraneous.
#x=2# or #x=-1#

Explanation:

Given #x^2+2 = x+4#

Subtract #(x+4)# from both sides to get into standard form
#color(white)("XXXX")##x^2-x-2=0#

Factor
#color(white)("XXXX")##(x-2)(x+1) = 0#

Either #(x-2)=0# or #(x+1)=0#
#rArr##color(white)("XXXX")##x=2# or #x=-1#

Checking #x=2# by substituting into #x^2+2# and #x+4#
#color(white)("XXXX")##x^2+2=(2)^2+2 = 6#
#color(white)("XXXX")##x+4=(2)+4 = 6#
#x^2+2=x+4# so #x=2# is not extraneous.

Similarly, checking #x=-1# by substituting into #x^2+2# and #x+4#
#color(white)("XXXX")##x^2+2=(-1)^2+2 = 3#
#color(white)("XXXX")##x+4=(-1)+4 = 3#
#x^2+2=x+4# so #x=-1# is not extraneous.