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

Aug 6, 2017

$x \in \emptyset$

#### 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

$2 x + 11 \ge 0 \implies x \ge - \frac{11}{2}$

Moreover, the square root of a positive number can only return a positive number, so you also know that

$\sqrt{2 x + 11} \ge 0$

This implies that

${\overbrace{6 + \sqrt{2 x + 11}}}^{\textcolor{b l u e}{\ge 6}} = - x$

So you can say that

$- x \ge 6 \implies x \le - 6 \mathmr{and} x \in \left(- \infty , - 6\right]$

However, you already know that you need to have $x \ge - \frac{11}{2}$, or $x \in \left[- \frac{11}{2} , + \infty\right)$.

Since

$\left(- \infty , - 6\right] \cap \left[- \frac{11}{2} , + \infty\right) = \emptyset$

you can say that the original equation has no real solution, or $x \in \emptyset$.