# 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