How do you solve #root4(2x)+root4(x+3)=0#?
1 Answer
This equation has no solutions.
Explanation:
Note that
So the only circumstances under which the left hand side of the given equation can be
#2x=0" "# and#" "x+3=0#
The first of these implies that
These are incompatible, so there are no real solutions.
Complex solutions?
Given:
#root(4)(2x) + root(4)(x+3) = 0#
Subtract
#root(4)(2x) = -root(4)(x+3)#
Raise both sides to the
#2x = x+3#
Subtract
#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.