How do you solve the equation and identify any extraneous solutions for #sqrt(x^2 + 2) = x + 4#?

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Answer 1 · 2015-06-19T21:24:58

I found #x=-7/4#

I would square both sides to get:
#x^2+2=(x+4)^2#
#cancel(x^2)+2=cancel(x^2)+8x+16#
#8x=-14#
#x=-14/8#
#x=-7/4#

Answer 2 · 2015-06-19T21:28:29

Square both sides to get:

#x^2+2 = x^2+8x+16#

Hence #x = -7/4#

Try squaring both sides to get:

#x^2+2 = (x+4)^2 = x^2+8x+16#

Subtract #x^2+16# from both sides to get:

#8x = -14#

Divide both sides by #8# to get:

#x = -14/8 = -7/4#

Check:

#sqrt(x^2+2) = sqrt(49/16+32/16) = sqrt(81/16) = 9/4#

#x+4 = -7/4 + 4 = -7/4 + 16/4 = 9/4#

graph{(y - sqrt(x^2+2))*(y - x - 4) = 0 [-9.42, 10.58, -1.8, 8.2]}