How do you solve the equation #6+sqrt(2x+11)=-x#?

1 Answer
Aug 6, 2017

Answer:

#x in O/#

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 #x >= -11/2#, or #x in [-11/2, + oo)#.

Since

#(-oo, - 6] nn [-11/2, + oo) = O/#

you can say that the original equation has no real solution, or #x in O/#.