How do you solve #(3x+4)^(1/3) = -5# and find any extraneous solutions?

1 Answer
Feb 25, 2017

Answer:

#x=-43#

Explanation:

#color(orange)"Reminder " x^(1/3)=root(3)x #

#rArr(3x+4)^(1/3)=root(3)(3x+4)#

To 'undo' the cube root, #color(blue)"cube both sides"# of the equation.

#rArr((3x+4)^(1/3))^3=(-5)^3#

#rArr3x+4=-125#

subtract 4 from both sides.

#3xcancel(+4)cancel(-4)=-125-4#

#rArr3x=-129#

divide both sides by 3

#(cancel(3) x)/cancel(3)=(-129)/3#

#rArrx=-43#

#color(blue)"As a check"#

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

#"left side "=((3xx-43)+4)^(1/3)=(-125)^(1/3)=-5#

#rArrx=-43" is the only solution"#

There are no extraneous solutions.