How do you solve #x^2 + 2 = x + 4# and find any extraneous solutions?

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Answer 1 · 2017-05-20T10:17:02

#x=2,-1#

#x^2+2=x+4#

#x^2+2-x-4=0#
Rearrange Equation

#x^2+2-x-4=0#
Collect like terms so #2-4=-2#

#x^2-2-x=0#
Plug values into the Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(-1)+-sqrt((-1)^2-4xx1xx(-2)))/(2xx1)#
Then solve
#x=(1+-sqrt(1+8))/(2)#

#x=(1+-sqrt(9))/(2)#

#x=(1+-3)/(2)#

#x=(1+3)/(2)#

#x=(4)/(2)#

#x=(1-3)/(2)#

#x=(-2)/(2)#

#x=2,-1#