How do you solve #sqrt(3x + 1) = -10#?

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Answer 1 · 2015-08-24T01:41:16

#x=33#

#sqrt(3x+1)=-10#

Square both sides of the equation.

(#sqrt(3x+1))^2=(-10)^2#

#3x+1=100#

Subtract #1# from both sides.

#3x=100-1##=#

#3x=99#

Divide both sides by #3#.

#x=99/3##=#

#x=33#

Check the answer by substituting #33# for #x#.

#sqrt(3x+1)=-10#

#sqrt(3*33+1)=-10#

#sqrt100=-10#

#(-10)^2=100#

#sqrt100=+-10#, so #33# is a valid number for #x#.