Solving and Graphing Radicals - Algebra 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 .

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

To get rid of the radical, we square both sides.

To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.

Square both sides to get rid of the radical.

Subtract on both sides.

Subtract two from both sides.

Simplify the equation.

Square both sides.

Add on both sides.

Subtract on both sides.

The answer is:

Square both sides to get rid of the radical.

Add on both sides.

Square both sides to get rid of the radical.

Divide on both sides.

Square both sides to get rid of the radical.

Multiply on both sides.

Add three on both sides.

The equation becomes:

Square both sides to eliminate the radical.

Add three on both sides.

Divide by three on both sides.

The answer is:

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 .

Subtract from both sides to group the radicals.

Square both sides.

Use the FOIL method to simplify the right side.

Combine like-terms.

Subtract one from both sides, and add on both sides.

The equation becomes:

Divide by two on both sides and distribute the terms inside the radical.

Square both sides.

Simplify the right side by FOIL method.

Subtract on both sides. This is the same as subtracting on both sides.

Subtract on both sides. The equation will become:

Multiply by four on both sides to eliminate the fractional denominator.

Use the quadratic equation to solve for the roots.

Simplify the radical and fraction.

Substitute the values of and back into the original equation, and only will satisfy both sides of the equation.

The answer is:

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.

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!