Quadratic Equations and Inequalities - Algebra 2

Combine like terms first.

Factor

The zeroes are 3 and 8 so a number line can be divided into 3 sections.

X<3 works, 3<x<8 does not work, and x>8 works

First we want to rewrite the quadratic in standard form:

Now we want to set it = 0 and factor and solve like normal.

Using the zero product property, both factors produce a zero:

So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, x = 0) and evaluate the original inequality.

Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.

Ergo, .

1. Rewrite the equation in standard form.

2. Set the equation equal to and solve by factoring.

So, and are our zeroes.

3. Test a point between your zeroes to find out if the solution interval is between them or on either side of them. (Try testing by plugging it into your original inequality.)

Because the above statement is true, the solution is the interval between and .

The discriminant of a quadratic equation in form is equal to . The given equation is not in that form however, so we must first multiply it out to get it into that form. We therefore obtain:

We therefore have , , and . Our discriminant is therefore:

The correct answer is therefore

Start by setting the inequality to zero and by solving for .

Now, plot these two points on to a number line.

Notice that these two numbers effectively divide up the number line into three regions:

, , and

Now, choose a number in each of these regions and put it back in the factored inequality to see which cases are true.

For , let

Since this is not less than , the solution to this inequality cannot lie in this region.

For , let .

Since this will make the inequality true, the solution can lie in this region.

Finally, for , let

Since this number is not less than zero, the solution cannot lie in this region.

Thus, the solution to this inequality is

First, set the inequality to zero and solve for .

Now, plot these two numbers on to a number line.

Notice how these numbers divide the number line into three regions:

Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.

For , let

Since this is not less than zero, the solution to the inequality cannot be found in this region.

For , let

Since this is less than zero, the solution is found in this region.

For , let

Since this is not less than zero, the solution is not found in this region.

Then, the solution for this inequality is

First, set the inequality to zero and solve for .

Now, plot these two numbers on to a number line.

Notice how these numbers divide the number line into three regions:

Now, you will choose a number from each of these regions to test to plug back into the inequality to see if the inequality holds true.

For , let

Since this solution is greater than or equal to , the solution can be found in this region.

For , let

Since this is less than or equal to , the solution cannot be found in this region.

For , let

Since this is greater than or equal to , the solution can be found in this region.

Because the solution can be found in every single region, the answer to this inequality is

Start by changing the less than sign to an equal sign and solve for .

Now, plot these two numbers on a number line.

Notice how the number line is divided into three regions:

Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.

For , let

Since this number is not less than zero, the solution cannot be found in this region.

For , let

Since this number is less than zero, the solution can be found in this region.

For let .

Since this number is not less than zero, the solution cannot be found in this region.

Because the solution is only negative in the interval , that must be the solution.

First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: . Now we know that the quadratic has zeros at and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this:

We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros and .

The only x-value satisfying the inequality is .

The value of would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.

(2x + 3)(x + 5)

First: 2x multiplied by x = 2x²

Outer: 2x multiplied by 5 = 10x

Inner: 3 multiplied by x = 3x

Lasts: 3 multiplied by 5 = 15

Put it all together: 2x² + 10x + 3x + 15

Simplify: 2x² + 13x + 15

First:

Outside:

Inside:

Last:

Make sure you keep track of negative signs when doing FOIL, especially when doing the Outer and Inner steps.

When the 2 terms differ only in their sign, the -term drops out from the final product.

In order to evaluate one needs to multiply the expression by itself using the laws of FOIL. In the foil method, one multiplies in the following order: first terms, outer terms, inner terms, and last terms.

Multiply terms by way of FOIL method.

Now multiply and simplify.

Use the FOIL method (First, Outer, Inner, Last):

Put all of these terms together:

Combine like terms:

Remember that FOIL stands for First, Outside, Inside, Last.

Multiply using FOIL and group like terms.

Expand the expression using the FOIL Method:

FIRST:

OUTER:

INNER:

LAST:

Therefore:

For this problem we need to use FOIL which stands for Firsts, Outters, Inners, Last. This describes the order of distributing and multiplying the binomials.

Firsts gets us

Outters gets us

Inners gets us

Lasts gets us

Rewriting these in an equation form we get:

Now we combine the like terms of -10x and 3x which results in: