Factoring a polynomial using a difference of squares

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Solve a system of equation with elimination

Mixture problem and system of equations

Solving a system of equations with substitution

Using the concept of system of equations in the context of uniform motion problems

Application of solving a system of equations

Solving linear inequalities

Solving double inequalities

Uniform motion equations

Calculating Confidence Intervals for Proportions

Solving mixture problems

Solving linear inequalities in word problems.

Calculating simple interest

Solving Linear equations involving word problems

Calculating standard deviation

Solving a linear equation

Solving a system of non-linear equations means that at least one of the equations is not linear. For example, if one equation has an exponent it may be a parabola or a circle. With this no shape that is not linear it involves slightly different expectations.

Solving a system of non-linear equations is similarÂ to solving a system with linear equations with one difference. The difference is that with nonlinear equations you can have more than one solution. What this means is that the lines that are the equations can intersect in more than one place. However, it is also possible they do not intersect. How many solutions depends on the lines involved.

For example, if one equation is a circle/parabola and the other is a line there can be 0-2 solutions. If one equation makes a circle and the other makes a parabola there can be up to 4 solutions. Two circles or two parabolas can make a multitude of solutions

The steps for solving a nonlinear system are the same. Therefore, in this post, we will demonstrate how to solve a system of non-linear equations using the substitution and elimination methods.

**Substitution**

The substitution method is when we plug one equation into the variable of the other equation. Below is our system of equations.

The first thing to notice is that the top equation would make a circle if you graphed it. That is why this is a non-linear system. To solve we take the second equation and substitute it for y. Below we find the values for x

Now we complete the system by finding the values for y.

THerefore are ordered pairs are (0, -3) and (1,0). These are the two points at which the equations intersect if you were to graph them.

**Elimination**

Elimination involves making the coefficients of one of the variables opposite so that when they are added together they cancel each other out. By removing one variable you can easily solve for the other. Below is the system of equation we want to solve.

The top equation makes a circle while the bottom one makes a parabola. This means that we can have as many as four solutions for this system. To solve this system we will multiply the bottom equation by -1. This will allow us to remove the x variable and then solve for y. Below are the steps.

Now we simply solve for y.

Now we can take these values for y to solve for x.

The order pairs are as follows

- (-2,0)
- (2,0)
- Â (âˆš3, -1)
- (-âˆš3, -1)

**Conclusion**

From this, you can see that non-linear equations can be solved using the same approaches. Understanding this is key to many other fields of math such as data science and machine learning.

A logarithm is the inverse of exponentiation. Depending on the situation one form is better than the other. This post will explore logarithms in greater detail.

**Converting Between Exponential and Logarithmic Form**

There are times when it is necessary to convert an expression from exponential to logarithmic and vice versa. Below is an example of who the expression is rearranged form logarithm to exponential.

The simplest way to explain I think is as follows

- for the logarithm, the exponent (y) and the base (a) are on opposite sides of the equal sign
- For the exponent form, the exponent (y) and base (a) are on the same side of the equal sign.

Here is an example using actual numbers

As you can see the exponent 3 and the base 2 are on opposite sides of the equal sign for the logarithmic form but er together for the exponential form.

When the base is e (Euler’s Number) it is known as a natural logarithmic function. e is the base rate growth of a continual process. The application of this is limitless. When the base is ten it is called a common logarithmic function.

**Logarithmic Model Example**

Below is an example of the application of logarithmic models

*Exposure to noise above 120 dB can cause immediate pain and damage long-term exposure can lead to hearing loss. What aris the decimal level of a tv with an intensity of 10^1 watts per square inch.Â *

First, we need the equation for calculating the decibel level.

Now we plug in the information into the word problem for I and solve

Our tv is dangerously loud and should include a warning message.Â We dropped the negative sign because you cannot have negative decibel level.

**Conclusion**

Logarithms are another way to express exponential information and vice versa. It is the situation that determines which to use and the process of concert an expression from one to another is rather simple. In terms of solving actual problems, it is a matter of plugging numbers into an equation and allowing the calculator to work that allows you to find the answer.

There are times when we want to understand growth that is not constant. An example of this would be the growth of a virus. As time goes by the virus growth rate increases more and more. Another example would be in the world of finance when we are dealing with interest.

In situations like the ones mentioned above, it is critical to understand the use and application of exponential models. This post will go through examples of the use of exponential models.

**Finance Example**

One common exponential model in finance is for compounded interest. The equation is as follows…

Below is a simple word problem that calls for this equation

*You invest $10,000 in a mutual fund to prepare for retirement. The interest rate is 5% compounded monthly, how much will be in the account when you plan to retire in 25 years.Â *

Below is what we now

- balance = ?
- principal = $10,000
- rate = 0.05
- years=Â 25
- times in year = 12 * 25 = 300

Now, we simply plug this information into the equatiom to get the answer.

The answer is shown above. The initial investment would grow to almost $35,000 dollars over 25 years.

**Continuous Growth**

In some fields, such as the life sciences, you want to now the growth of a virus or bacteria. Unlike in finance where the balance grows several times a year,Â a bacteria is growing continuously. This leads to a slightly different exponential model as shown below.

e is an irrational number that serves as the base. With this information, we can address the problem below

*A student starts their experiment with 10 bacteria. He knows the bacteria grow 100% every hour. He will come back and check in 12 hours. How many bacteria will he find?*

Here is what we know

- final size =?
- initial size = 10
- rate = 1/hour
- time = 12

We plug this into the equation to get the answer

As you can see, the growth of the bacteria is almost incomprehensible in such a short time. This is the power of exponential growth.

**Conclusion**

Exponential models provide another way to find answers to questions people have. Whether the growth is over a certain number of times or continuously the model can be adjusted to deal with either of this situations.

One method for solving quadratic equations is called completing the square. This approach is a little confusing but we will try to work through it together in this post.

**What is Completing a Square**

Completing the square is used when your quadratic equation is not a perfect square. Below is an example of a perfect square quartic formula =. The first is in the standard quadratic form and the second is after it has been simplified.

However, not all equations are this easy, consider the example below.

There is no quick way to factor this as there is no perfect square. We have to use something called the binomial square pattern.

This is where it gets confusing but essential what the binomial square pattern is saying is that if you want to find the third term (b squared) you must take the second term and multiple it by1/2. We multiplied by 1/2 because this is the reciprocal of multiplying by 2 as shown in the equation. Lastly, we square this value. Below is the application of what we just learned from our problem equation.

By taking the second term, multiplying by 1/2 and squaring it we were able to create the trinomial we needed to create the perfect square. By doing this we also solved for x if this was a full equation.

**Examples**

When using the completing the square approach with a quadratic formula there are some additional steps. We will work through an example below

We are missing the third term and we need to find this first. Our second term is 8 so we will plug this in to find the third term

We take this number 16 and add it to both sides which is a rule whenever manipulating an equation. Therefore, we get the following.

We can now factor the left side as shown below.

To remove the square we need to square root both sides. In other words, we are employing the use of the square root property.

This leads to our two answers as shown below

There are variations of this but they all involve just moving some numbers around before the steps shown here. As such, there is not much need to discuss them.

**Conclusion**

Completing the square provides a strategy for dealing with quadratic formulas that do not have a perfect square. Success with this technique requires identifying the terms you know and do not know and taking the appropriate steps to calculate the third term for the trinomial.

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