How do you solve #\sqrt { 54+ 3x } = x#?

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Answer 1 · 2017-03-29T16:44:04

#x=9 or x=-6#

Square both sides

#54+3x=x^2#

Take #54# and #3x# from both sides

#x^2-3x-54=0#

Then we Differentiate

#(x-9)(x+6)=0#

#x=9 or x=-6#

Answer 2 · 2017-03-29T16:52:07

#x_1=9#; #x_2=-6#

Squaring both sides, we can get the left side of the equation out of the square root.
#54+3x=x^2#
#x^2-3x-54=0#
#(x-9)(x+6)=0#

Hence, the roots of #x# are:
#x_1=9# and #x_2=-6#

Answer 3 · 2017-03-29T17:12:34

#color(red)(x=9 and x=-6#

#sqrt(54+3x)=x#

#:.(sqrt(54+3x))^2=x^2#

#:.sqrt(54+3x)*sqrt(54+3x)=x^2#

Note:

#color(red)(sqrta*sqrta=a#

#:.x^2=54+3x#

#:.x^2-3x-54=0#

#:.(x-9)(x+6)=0#

#:.color(red)(x=9 and x=-6#