Solving and Graphing Radicals - Algebra II

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.

The above graph is not a function. In order to be a function, the graph must pass the vertical line test. If you look at x=2, you can see that x=2 is a value on both the green graph and the purple graph. Therefore it fails to pass the vertical line test and is not a function.

This graph is a function. We call this type of function a piecewise (or step) function. This type of function is defined in a certain way for certain x values, and then in a different way for different x values. However, it still passes the vertical line test, so it is a function!

Firstly, this is a function as it passes the vertical line test.

To find the domain, begin at the far left of the graph, starting with the red portion. You can trace along it continuously until you get to the point (-5,-5). At this point, it jumps upward to the green graph. While the y values jump, there is no gap in the x values. You can trace from left to right along the green graph until you get to x=2. Then, the graph jumps up to the purple values. Once again, there is no gap in the x values. The purple graph is continuous and goes onwards to positive infinity. Therefore, the domain of this graph is all real numbers or .

Next, let's start again at the bottom of the graph (in red), this time looking at the y values. We begin at negative infinity and can trace continuously until we get to the point (-5,-5). There is a discontinuity between the red and green graphs' y values. We use the symbol ("union") to connect the disconnected parts of the graph. The green graph picks up at -1.71 and the y values continue upward until they get to 1.256. Once again, there is a discontinuity in the range, so we again use the union symbol. The y values pick back up at the point (2,4) and continue upwards to infinity. Therefore the range of this function is .

The graph of is

Keep in mind that you cannot put a negative number under a square root and get a real number answer. If we tried to plug in -2 for x, it would be undefined. That is why the graph only shows positive values of x.

The other graphs pictured below are , , and .

The graph of is

.

The other graphs shown are those of , , and .

The graph of is

The other graphs pictured are , , and .

This graph crosses the x axis only once and the y axis only once, and these happen to coincide at the point (0,0). Therefore the x-intercept is at x=0 and the y-intercept is at y=0. The graph will go on infinitely to both the left and the right. Additionally (and despite the fact that it won't happen as quickly), the graph also will go infinitely high in both its positive and negative y values. Therefore, both the domain and range of this graph are .

This graph intersects the x axis only once and the y axis only once, and these happen to coincide at the point (0,0). Therefore the x-intercept is at x=0 and the y-intercept is at y=0. The graph will go on infinitely to the right. However, the leftmost point is (0,0.) Likewise, the graph will go infinitely high, but its lowest point is also (0,0). Therefore, its domain is and its range is also .

One way to solve this equation is to substitute for and, subsequently, for :

Solve the resulting quadratic equation by factoring the expression:

Set each linear binomial to sero and solve:

or

Substitute back:

- this is not possible.

- this is the only solution.

None of the responses state that is the only solution.

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.

We can simplify the fraction:

Plugging this into the equation leaves us with:

Note: Because they are like terms, we can add them.

In order to solve this equation, we need to know that

How? Because of these two facts:

1. 
2. Power rule of exponents: when we raise a power to a power, we need to mulitply the exponents:

With this in mind, we can solve the equation:

In order to eliminate the radical, we have to square it. What we do on one side, we must do on the other.

In order to solve this equation, we need to know that

Note: This is due to the power rule of exponents.

With this in mind, we can solve the equation:

In order to get rid of the radical we square it. Remember what we do on one side, we must do on the other.

To solve, perform inverse opperations, keeping in mind order of opperations:

first, square both sides

subtract 1

divide by 2

To solve, perform inverse opperations, keeping in mind order of opperations:

take the square root of both sides

subtract 19 from both sides

square both sides

To solve, use inverse opperations keeping in mind order of opperations:

divide both sides by 5

square both sides

add 12 to both sides

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 .