How do you solve $root4(2x)+root4(x+3)=0$ ?

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Answer 1 · 2017-04-29T13:58:51

This equation has no solutions.

Note that $root(4)(t) >= 0$ for all $t >= 0$ and has no real value when $t < 0$ .

So the only circumstances under which the left hand side of the given equation can be $0$ are if both:

$2x=0" "$ and $" "x+3=0$

The first of these implies that $x=0$ and the second that $x=-3$ .

These are incompatible, so there are no real solutions.

$color(white)()$
Complex solutions?

Given:

$root(4)(2x) + root(4)(x+3) = 0$

Subtract $root(4)(x+3)$ from both sides to get:

$root(4)(2x) = -root(4)(x+3)$

Raise both sides to the $4$ th power (noting that this is where we might introduce extraneous solutions) to get:

$2x = x+3$

Subtract $3$ from both sides to find:

$x = 3$

But...

$root(4)(2(color(blue)(3)))+root(4)(color(blue)(3)+3) = root(4)(6)+root(4)(6) != 0$

So there are no solutions at all.

How do you solve root4(2x)+root4(x+3)=0root4(2x)+root4(x+3)=0?