Question

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

Answer 1

`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.