How do you solve #sqrt(4x+8)+8 = 1#?

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Answer 1 · 2018-05-06T11:06:49

#x = 10.25#

#sqrt(4x+8) = 1 - 8#
#sqrt(4x+8) = -7#
#(sqrt(4x+8))^2 = (-7)^2#
#4x+8 = 49#
#4x = 41#
#(4x) / 4= 41 / 4#
#x = 41 / 4#
#x = 10.25#

Answer 2 · 2018-05-06T12:26:31

#"no solution"#

#"isolate "sqrt(4x+8)" by subtracting 8 from both sides"#

#rArrsqrt(4x+8)=-7#

#sqrt(4x+8)>=0to x inRR#

#rArr"there is no solution"#