The square-root sign always denotes the "principal" root; in other words, when a Real number has two square roots, the square-root symbol stands for the positive root.

The square root in this exercise equals __negative__ seven. Since that is impossible, there is no solution for x in this equation.

I suspect that your daughter's teacher might have included this exercise to show what happens when we don't check our answers. Checking solutions is important with these types of equations.

Solving equations with radicals involves squaring both sides of the equation, and this process sometimes leads to false solutions.

Here's how the solution process goes.

\(\displaystyle \sqrt{6x} \;+\; 9 \;=\; 2\)

Isolate the radical.

\(\displaystyle \sqrt{6x} \;=\; -7\)

Square both sides.

\(\displaystyle 6x \;=\; 49\)

Solve for x.

\(\displaystyle x \;=\; \frac{49}{6}\)

Now, let's check this candidate by substituting it for x in the original equation.

\(\displaystyle \sqrt{6 \cdot \frac{49}{6}} \;+\; 9 \;=\; 2\)

\(\displaystyle \sqrt{49} \;+\; 9 \;=\; 2\)

\(\displaystyle 7 \;+\; 9 \;=\; 2\)

\(\displaystyle 16 \;=\; 2\)

Our candidate results in a false statement, so there is no solution.