Rewrite in standard form and factor:

The zeroes of the polynomial are therefore , so we test one value in each of three intervals , , and  to determine which ones are included in the solution set.

:

Test :

False;  is not in the solution set.

:

Test 

True;  is in the solution set

:

Test :

False;  is not in the solution set.

Since the inequality symbol is , the boundary points are not included. The solution set is the interval .

The first step of questions like this is to get the quadratic in its standard form. So we move the  over to the left side of the inequality:

This quadratic can easily be factored as. So now we can write this in the form

and look at each of the factors individually. Recall that a negative number times a negative is a positive number. Therefore the boundaries of our solution interval is going to be when both of these factors are negative.  is negative whenever , and is negative whenever . Since , one of our boundaries will be . Remember that this will be an open interval since it is less than, not less than or equal to.

Our other boundary will be the other point when the product of the factors becomes positive. Remember that is positive when , so our other boundary is . So the solution interval we arrive at is 

When asked to solve for x we need to isolate x on one side of the equation.

To do this our first step is to subtract 7 from both sides.

From here, we divide by 4 to solve for x.

When asked to solve for y we need to isolate the variable on one side and the constants on the other side.

To do this we first add 9 to both sides.

From here, we divide by -12 to solve for y.

The region  contains only  values which are greater than or equal to those on the line , so its  values are .

Also, the region contains only  values which are less than or equal to those on the line , so its  values are .

The region  contains only  values which are greater than or equal to those on the line , so its  values are .

Similarly, the region contains only  values which are less than or equal to those on the line , so its  values are .

To begin, we analyze the equation given: the base equation,  is shifted left one unit and vertically stretched by a factor of 2. The graph of the equation  is:

To solve the inequality, we need to take a test point and plug it in to see if it matches the inequality. The only points that cannot be used are those directly on our parabola, so let's use the origin . If plugging this point in makes the inequality true, then we shade the area containing that point (in this case, outside the parabola); if it makes the inequality untrue, then the opposite side is shaded (in this case, the inside of the parabola). Plugging the numbers in shows:

Simplified as:

Which is not true, so the area inside of the parabola should be shaded, resulting in the following graph:

The equation for a circle centered at point  with radius  is . Our circle is centered at  with , and we are interested in points that lie along or inside of the circle. This means the left-hand side must be less than or equal to the right-hand side of the equation. We are left with  or 

This is a graph of a circle with radius of 5 and a center of (1,1). The center of the circle is not on the edge of the circle, so that is the correct answer. All other points are exactly 5 units away from the circle's center, making them a part of the circle.

The center of the circle is , so plugging those values in for x and y yields the response that 0 is greater than or equal to 25.

Since plugging in the center values gives us a false statement we know that our center does not satisfy the inequality.