Which equation shows a valid, practical step in solving

Question

asked Apr 5, 2020 175k views

1 Answer

Don Brody · Answer 1 · 2020-04-11T06:10:40+0000

For this case we have the following equation:

$\sqrt [4] {2x-8} + \sqrt [4] {2x + 8} = 0$

If we subtract both sides of the equation
$\sqrt [4] {2x + 8}$ we have:

$\sqrt [4] {2x-8} = - \sqrt [4] {2x + 8}$

To eliminate the radical we raise both sides of the equation to the fourth power:

$(\sqrt [4] {2x-8}) ^ 4 = (- \sqrt [4] {2x + 8}) ^ 4$

Answer:

Option D