How do you solve $root3(3x - 8) = 1$ and find any extraneous solutions?

Question

1975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-21T22:26:02

$x=3$

There are no extraneous solutions.

$root3(3x - 8) = 1$

$(root3(3x - 8))^3 = 1^3$

$3x-8=1$

$3x=9$

$x=3$

test for "extraneous solutions":

$root3(3x - 8) = 1$

$root3(3(3) - 8) = 1$

$root3(9- 8) = 1$

$root3(1) = 1$

$1=1$ Checks.

How do you solve root3(3x - 8) = 1 root3(3x - 8) = 1 and find any extraneous solutions?