How do you solve #(3x+4)^(1/3) = -5# and find any extraneous solutions?
1 Answer
Feb 25, 2017
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.