How do you solve the equation #6+sqrt(2x+11)=-x#?
1 Answer
Aug 6, 2017
Explanation:
For starters, you know that when working with real numbers, you can only take the square root of a positive number, so you can say that you must have
#2x + 11 >= 0 implies x >= -11/2#
Moreover, the square root of a positive number can only return a positive number, so you also know that
#sqrt(2x + 11) >= 0#
This implies that
#overbrace(6 + sqrt(2x + 11))^(color(blue)(>= 6)) = -x#
So you can say that
#-x >= 6 implies x <= -6 or x in (-oo, -6]#
However, you already know that you need to have
Since
#(-oo, - 6] nn [-11/2, + oo) = O/#
you can say that the original equation has no real solution, or