mathwithhotti: Recent Episodes
Tom Hottinger
In this example we use the rational root theorem and synthetic substitution to find the zeros and factors of a polynomial to the third degree.
In this episode, there are 2 examples of verifying trig identities that I would rate on the more difficult side. More advanced strategies are needed such as factoring and finding a common denominator. While that sounds easy, it is often difficult to see when verifying trig identities.
In this episode, we look at verifying trig identities. There are 3 examples that I would rate on the easier side. You need to start somewhere.
This podcast is not about math. This is just a simple explanation of what a podcast is. I have created this podcast to help my students learn more about podcasts. They will eventually help me make more math podcasts. I am very much so looking forward to this project.
Here is another example how to use substitution to solve a quadratic equation. Now I know that there are other ways to solve this particular example (squaring both sides), but I wanted to show that substitution is a good way to solve a quadratic equation.
Sometimes we will encounter a problem that just looks ugly. In other words, it is one of those problems we just would rather skip. However, as you will see in this problem, substituting a quantity for a single variable really does make the problem much more manageable.
Here is another example of graphing a quadratic equation that is in vertex form. In this example, a is negative which causes the graph to open down and frown:-(
Graphing a quadratic equation is not a difficult task. Here are the main things to find: the vertex, the axis of symmetry, the y-intercept, and the x-intercepts. Use the fact that a parabola is symmetrical also makes it easier for us to find additional points.
Here is a bit of fun I had with one of my Alg II classes reviewing the quadratic formula.
A great way to solve a quadratic equation is by using the quadratic formula. It is straight forward and user friendly in my opinion. Many students prefer using the quadratic formula over the other leading brands of solving a quadratic equation.
In this next example, you will see just how great completing the square really is. At the end, you might recognize the solution.
Some students do not like to use the process known as completing the square. However, I will show you just how easy it really is to complete the square.
In this example, we are going to solve a quadratic equation by completing the square. In this example, the coefficient of our quadratic term is not 1.
In this example, we are going to solve a quadratic equation by completing the square. In this example, the coefficient of our quadratic term is 1.
In our next example, we are going to solve an equation that contains two radicals. No matter what we do, we must always try to isolate the radical before we start to solve the equation. It does not make a difference which radical you isolate, but isolate one of them. Once the radical is isolated, we can square both sides of our equation. Combine like terms. Notice we still have a radical left over. Isolate that radical now. Once that radical is isolated, square both sides of the equation again. Then we can solve the resulting quadratic equation.
In like our next like example, like we are like going to like solve a radical equation. Oh, sorry about that. Let's try that again.
In our next example, we are going to solve an equation that contains a radical. No matter what we do, we must always try to isolate the radical before we start to solve the equation. Once the radical is isolated, we can square both sides of our equation. Solve the resulting equation whether it is a simple linear equation or a quadratic equation.
Some think that fractions are complex enough. But when there is a fraction within a fraction, it really gets complex. To make the problem easier, multiply both the numberator and denominator by the least common multiple to get rid of the fractions within the fractions. You might need to factor and reduce when you are done.
In this example, we solve an equation with fractions but with varibles in the denominator. If you do not like to work with fractions, find the least common multiple of the denominator. Then multiply each term by the least common multiple and voila the fractions are gone and you can solve the equation that is left.
We will need to check to see if our answer is extraneous. Plug the solutions into the original equation and check to see if zero is produced in the denominator. If so, then the solution is extraneous. Only put solutions that are not extraneous in your solution set.
In this example, we solve an equation with fractions but with variables in the denominator. If you do not like to work with fractions, find the least common multiple of the denominator. Then multiply each term by the least common multiple and voila the fractions are gone and you can solve the equation that is left.
We do need to check our solution as it could be extraneous, a solution that does not work in the original problem. Why? Well, we know that in a fraction, the denominator can not be zero. We cannot divide by zero as there is no reciprocal for zero.