Add 5 on both sides.  This is the same as adding  on both sides.

To eliminate the radical, we must square both sides of the equation.

The answer is:  

Multiply the denominator on both sides to eliminate the denominator.

Simplify both sides.

Subtract 12 from both sides.

The equation becomes:

Divide by 4 on both sides.

Raise both sides by the power of four to eliminate the radical.

Multiply by one half on both sides to isolate .

The answer is:  

Subtract five from both sides of the equation.

Divide by negative three on both sides.

Square both sides.

Simplify both sides.

Multiply by one-ninth on both sides.

The answer is:  

Cross multiply both sides.

Square both sides in order to cancel the radicals.

Divide by 2 on both sides.

Square root both sides.

Only  will satisfy the original equation.

The answer is:  

We will begin by adding 5 to both sides of the equation. 

Next, let's square both sides to eliminate the radical.

Finally, we can solve this like a simple two-step equation. Subtract 3 from both sides of the equation.

Now, divide each side by 2. 

Finally, check the solution to make sure that it results in a true statement.

When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals.  So a good first step is to add  and subtract  from both sides so that you get:

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

Then when you subtract  from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

This factors to:

Which would seem to yield solutions of  and .  However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions.  Here if you plug in  the math holds:  because , and .

But if you plug in , you'll see that the original equation is not satisfied:  because  does not equal .  Therefore the only proper answer is .

To solve, first square both sides:

squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply :

This is a quadratic, we just need to combine like terms and get it equal to 0

now we can solve using the quadratic formula:

This gives us 2 potential answers:

and

Since the expression under the radical cannot be negative,

.

Solve for x:

This is the domain, or possible  values, for the function.

Because the term  inside the radical in the numerator has to be greater than or equal to zero, there is a restriction on our domain; to describe this restriction we solve the inequality for :

Add  to both sides, and the resulting inequality is:

     or   

Next, we consider the denominator, a quadratic; we know that we cannot divide by zero, so we must find x-values that make the denominator equal zero, and exclude them from our domain:

We factor an  out of both  and  :

Finding the zeroes of our expression leaves us with:

Therefore, there are 3 restrictions on the domain of the function:

The graph of  is: 

This graph is translated 2 units left and 4 units down from the graph of , which looks like:

The number 4 and its associated negative sign, which fall outside of the square root, tell us that we need to shift the graph down four units. If you take what's under the square root, set it equal to 0, then solve, you'll get x=-2, which you can interpret as being shifted left two units. 

Each of the graphs below that are incorrect either improperly shift the graph right (instead of left), up (instead of down), or both right and up.