## Solving Radical Equations: Algebra 2/Trig.

Today we’re back with Algebra 2, this time solving for radical equations!  Did you say “Radical Equations?” As in wild and crazy equations? No, not exactly, radicals in math are used to take the square root, cubed root, or whatever root of a number.

## Example #1:

Radicals are actually pretty cool because we can write them a couple of different ways and they all mean the same thing! Check it out below:Still not sure of their coolness? Let’s see what they look like with actual numbers:
Example: Solve the following algebraic equation below for the missing variable (aka, solve for x).Explanation:

How do I answer this question?

The question wants us to solve for x using our knowledge of radicals and algebra. You can also check out how to solve this question on Youtube here!

How do we do this?

Step 1: We start solving this radical equation like any other algebraic problem: by getting x alone. We can do this easily by subtracting 7 and then dividing out 5.

Step 2: Now, to get rid of that pesky radical, we need to square the entire radical.  Remember, whatever we do to one side of the equation, we must also do to the other side of the equation, therefore, we also square 14 on the other side of the equal sign. *This gets rid of our radical and allows us to solve for x algebraically as normal!

What happens when there is a cubed root though!?!?

When dividing polynomials with different value roots, raise the entire radical to that same power of root to cancel it out:Remember, we know radicals can also be written as fractions:

Therefore we also know that if we raise the entire radical expression to the same power of the root, the two exponents will cancel each other out:

## Example #2:

Want more practice? Try solving radical equations in the next few examples on your own.

## Solutions:

Looking to brush up on how to solve absolute value equations? Check out the post here! Did I miss anything?  Don’t let any questions go unchecked and let me know in the comments! Happy calculating! 🙂

Don’t forget to check out the latest free videos and posts with MathSux and subscribe!

## Dividing Polynomials: Algebra 2/Trig.

Greeting math peeps! In this post we are going to go over dividing polynomials! At some point, you may need to know how to answer these types of questions. The cool thing about dividing polynomials is that it’s the same long division you did way back in grade school (except now with a lot of x). Ok, let’s get to it and check out the question below:

Also, if you haven’t done so, check out the video related that corresponds to this problem on Youtube below! 🙂

Explanation:

How do I answer this question?

The question wants us to divide polynomials by using long division.

How do we do this?

Step 1: First we set up a good ole’ division problem with the divisor, dividend, and quotient to solve.

Step 2: Now we use long division like we used to back in the day! If you have any confusion about this please check out the video in this post.

## What if there’s a Remainder?

What happens when there is a remainder though!?!? When dividing polynomials with a remainder in the quotient, the answer is found and checked in a very similar way! Check it out in the example below:

Notice we represented the remainder by adding  to our quotient! We just put the remainder over the divisor to represent this extra bit of solution.

Want more practice? Try solving these next few examples on your own.

## Solutions:

If you’re looking for more on dividing polynomials, check out this post on synthetic division and finding zeros here!

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating! 🙂

## Binomial Cubic Expansion: Algebra 2/Trig.

Hey math friends! In this post, we are going to go over Binomial Cubic Expansion by going step by step! We’ll start by reviewing an old Regents question. Then, to truly master the topic, try the practice problems at the end of this post on your own! And, if you still have questions, don’t hesitate to watch the video or comment below. Thanks for stopping by and happy calculating! 🙂

Also, if you’re looking for more on Binomial Cubic Expansion, check out this post here!

## What are Cubed Binomials?

Binomials are two-termed expressions, and now we are cubing them with a triple exponent! See how to tackle these types of problems with the example below:

How do I answer this question?

We need to do an algebraic proof to see if (a+b)3=a3+b3.

How do we do this?

We set each expression equal to one another, and try to get one side to look like the other by using FOIL and distributing. In this case, we will be expanding (a+b)3 to equal (a+b)(a+b)(a+b).

Extra Tip! Notice that we used something called FOIL to combine (a+b)(a+b).  But what does that even mean? FOIL is an acronym for multiplying the two terms together.  It’s a way to remember to distribute each term to one another.  Take a look below:

Add and combine all like terms together and we get !

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## Rational Exponents: Algebra 2/Trig.

Hi everyone and welcome to MathSux! In this post we are going to break down and solve rational exponents. The words may sound like a mouthful, but all rational exponents are, are fractions as exponents. So instead of having x raised to the second power, such as x2, we might have x raised to the one-half power, such as x(1/2). Let’s try an example taken straight from the NYS Regents below. Also, if you have any questions don’t hesitate to comment below or check out the video posted here. Happy calculating! 🙂

How do I answer this question?

The questions want us to simplify the rational exponents into something we can understand.

How do we do this?

We are going to convert the insane looking rational exponents into radical and solve/see if we can simplify further.

Reminder!

A radical can be converted into a rational exponent and vice versa. Not sure what that means? It’s ok! Take a gander at the examples below and look for a pattern:

Think you’re ready to take on our original problem? #Letsdothis

## Solutions:

Still got questions?  Don’t hesitate to comment below for anything that still isn’t clear! Looking to review how to solve radical equations? Check out this post here! 🙂

Also, don’t forget to follow MathSux fopr FREE math videos, lessons, practice questions and more every week!

## How to Factor Trig Functions

Howdy math friends! In this post, we are going to learn how to factor trig functions algebraically.  This will combine our knowledge of algebra and trigonometry into one beautiful type of question! For more on trigonometric equations and the unit circle, check out this post here.

Solving trigonometric equations is very similar to solving the ho-hum everyday equations you solved all the way back in algebra 1, but this time we are solving for an angle value. If you need to refresh yourself on how to find x the old fashioned way via factoring be sure to check out this post here!

Solving trigonometric equations, sounds complicated? Well, you are correct, that does sound complicated. Is it complicated? Hopefully, you won’t find it that way after you’ve seen an example. We are going to find the value of x (value of the unknown angle) step by step in the following question:

## How do I answer this question?

The questions want us to solve for x, the unknown value of the angle.

Step 1: Our first step for solving trigonometric equations is to pretend this is any other algebraic equation and we want to solve for x (Note: this is the key to solving any trig equation). Thinking this way, allows us to move radical 2 to the other side, by adding it to both sides of the equation.

Step 2: Remember that secant is the inverse trigonometric function of cosine? Therefore we know that, secx is equal to 1/cosx and we can re-write the equation below, making the trigonometric function look a lot less complicated already!

Step 3: Now we need a value for x. This is where I turn to my handy dandy special triangles. There are two main special triangles in trignometry we can use to solve our equation, the 45 45 90 triangle and the 30 60 90 triangle. In this case, we are going to use the 45 45 90 special triangle, because its proportions include a value of radical 2, which is exactly what our equation has!

*Note- In this case we can use our special triangles, but in other examples, we may also need to use our knowledge of trig identities to re-write the original trigonometric function.

Below we can see the proportions that the 45 45 90 special triangle brings us. When we apply the basic rules of trigonometry and SOH CAH TOA, we see that we can find the value of our missing angle x, which is 45º.

*Another way we could have gotten 45º without using special triangles is by plugging in cos-1 (1/rad2) into our calculator, this would also give us the solution of 45º!

Step 4: We have a value for cos45º=1/rad 2, but remember we need its inverse and must flip the fraction, to find the value of our angle of the inverse of cosine, secx, which flipped, is going to be 45º.

***But wait! We are not done yet! We have found only one part to our solution, x=45º, which is great! But we need to ask ourselves, are there any other angle measures that could potentially give is the right solution as well!? And thats where we need to consult the unit circle!

Step 5: Using the unit circle below, we can identify what other values can work for our original trigonometric function. Based on the unit circle below, we can identify which trig functions (sin, cos, tan) are positive in which quadrants, positive and negative angles (positive counterclockwise, negative clockwise).

Notice that cosine is positive in Quadrants I and IV. That means there are two values that x can be (one in each quadrant). We already have x=45º from Quadrant i. In order to get the other value in Quadrant IV, we must subtract 360º-45º=315º using our reference angle (360º-θ) and giving us our other value for x, 315º, completing our answer!

The above example is similar to a linear equation, let’s look at another type of trigonometric equation that is more similar to quadratic equations.

## Factoring Trig Functions Example: Quadratic Equation

Step 1: First let’s move everything onto one side of the equation.  We can start by multiplying tanθ to both sides and distributing, multiplying (2tanθ-7)(tanθ). Then we can subtract the 4 from both sides.  Remember to solve trigonometric equations, we always want to treat them like a normal algebraic function.

Step 2: Now that we have something resembling a quadratic equation, we have something to work with! There are two main ways to solve for our unknown angle via factoring for this example, (1) The Quadratic Formula, or, (2) Factor by Grouping.  In this case, we are going to be using factor by grouping to find the value of our angle, and the solution to our problem, but the quadratic formula should work as well!

Factor by Grouping: First find the product, sum and factors of the equation.

Now that we have our factors, we can re-write our equation, converting -7tanθ into -8tanθ + tanθ.

Next, we split the equation down the middle and take the GCF of each side.

Finally, we can combine the values outside our parenthesis 2tanθ+1 and the matching values found within our parenthesis and tanθ-4.

Step 3: Now we can use our equation and solve for each angle value, by setting up each group found in the parenthesis equal to zero, just like we would when finding the value of x in a normal algebraic equation.

Based on finding our two angles for θ, -27º and 76º, we are able to find the angles that apply to our specific trigonometric equation listed below, based on where tanθ is positive and negative in each quadrant of the unit circle. In this case, the angle value of -27º, we look for the angle values where tangent is negative which are in quadrant ii and quadrant IV, and apply our reference angles. On the other hand, the angle value of 76º, is positive, therefore, we look for angles where tangent is positive, in quadrant i and quadrant iii, and apply our reference angles to get the correct value of each angle.

*Note! In the above two examples, we look at a cosine function and a tangent function, but the same exact methods apply to a sine function!

Think you are ready to solve trigonometric equations on your own? Check out the practice questions and solution to each below!

## Practice Questions:

Find the nearest degree to all values of θ or x in the interval 0º <  θ < 360º and 0º <  x < 360º:

When you are ready check each solution below!

## Solution:

Does the solution to each trig function make sense? Great! 🙂 Do you think it is clear as mud to solve trigonometric equations? Alas, I have failed. But I have not given up (and neither should you). Ask more questions, look for the spots where you got lost, do more research and never give up! 🙂

Hopefully, you enjoyed my short motivational speech. For more encouraging words and math, check out MathSux on the following websites! Sign up for FREE math videos, lessons, practice questions, and more. Thanks for stopping by and happy calculating! 🙂

## Related Trigonometry Posts:

The Unit Circle

Basic Right Triangle Trigonometric Ratios (SOH CAH TOA)

4545 90 Special Triangles

30 60 90 Special Triangles

Graphing Trigonometric Functions

Trig Identities

Transforming Trig Functions

Law of Cosines

Law of Sines

## Factor by Grouping: Algebra 2/Trig.

Hey math friends! In this post, we are going to go over Factor by Grouping, one of the many methods for factoring a quadratic equation.  There are so many methods to factor quadratic equations, but this is a great choice, for when a is greater than 1. Also, if you need to review different types of factoring methods, just check out this link here.  Stay curious and happy calculating! 🙂 Before we go any further, let’s just take a quick look at what a quadratic equation is: Usually, we can just find the products, the sum, re-write the equation, solve for x, and be on our merry way.  But if you notice, there is something special about the question below. The coefficient “a” is greater than 1/. This is where factor by grouping comes in handy! Now that we know why and when we need to factor by grouping lets take a look at our Example:

Factor By Grouping: Hard to solve? No.  Hard to remember? It can be, just remember to practice, practice practice! Also, if you are in need of a review of other methods of factoring quadratic equations, click this link here.

Want more Mathsux?  Don’t forget to check out our Youtube channel and more below! And if you have any questions, please don’t hesitate to comment. Happy Calculating! 🙂

Looking for more on Quadratic Equations and Functions? Check out the following Related posts!

Factoring Review

What is the Discriminant?

Completing the Square

4 Ways to Factor a Trinomial

Is it a Function?

Imaginary and Complex Numbers

Quadratic Equations with 2 Imaginary Solutions

Focus and Directrix of a Parabola

## Recursive Formula: Algebra

Howdy math peeps! In this post, we are going to go over the recursive formula step by step by reviewing an old Regent’s question. These things may look weird, confusing, and like a “what am I doing?” moment, but trust me they are not so bad! We are going to take a look at the Regents question below and then find the correct recursive formula that follows the given sequence by going through each answer choice. Before we begin to answer our questions though, we will first define and break down what a recursive sequence is. Also, be sure to check out the video at the end of this post for more examples and further explanation. Happy calculating!

Before we dive into the solution to this question, let’s first look at a recursive formula example and define what they are in the first place!

## What is the Recursive Formula?

A Recursive Formula is a formula that forms a sequence based on the previous term value. All this means is that it uses a formula to form a sequence-based pattern. Recursive formulas can take the shape of different types of sequences, including arithmetic sequences (sequence based on adding/subtracting numbers) or geometric sequences (sequence based on multiplying/dividing numbers). Check out the example below:

a1=2 , an+1=an+4

## How do You Solve a Recursive Equation?

When solving a recursive equation, we are always given the first term and a formula. We start by using the first given term of our sequence, usually represented as a1, and plug it into the given formula to find the value of the second term of the sequence. Then we take the value of the previous term (the second term in our sequence) and plug it into our formula again, to find the third value of our sequence…. and the pattern continues! The solutions we get from each step up forms a sequence. If this sounds confusing, don’t worry because we are going to look at an example! Take a look at the recursive formula below:

a1=2 , an+1=an+4

Now let’s take another look at our recursive formula, this time breaking down what each part means:

a1 always represents the first given term, which in this case is a1=2. Next, we plug in 2 for an into our formula an+4 to get (2)+4=6. This gives us our second term in the sequence, which is 6. Next, we plug in 6 into the formula to get (6)+4=10, which is the value of the third term in the sequence. And we continue the pattern, always taking the value of the preceding term! In this case, I just found the first few terms below, but the recursive formula can continue infinitely! Again , if this sounds confusing, please take look at the pattern below:

Terms of the Sequence: 2, 6, 10, 14, 18….

Notice all the terms of the sequence are circled in pink, forming a sequence of, 2, 6, 10, 14, 18!? This is what the recursive formula produces!

## Now back to our Original Question:

Q: What recursively-defined function represents the sequence 3, 7, 15, 31,

(1) f(1) = 3, f(n + 1) = 28 (n) +3

(2) f(1) = 3, f (m + 1) = 28(0) – 1

(3) f(1) = 3, f(n + 1) = 28 (n) +1

(4) f(1) = 3, f (n + 1) = 38 (n) -2

How do I answer this question?

At first glance, all of these answer choices may look exactly the same as there are many recursive formulas to choose from. The first thing we need to do is to identify how each answer choice is different. Notice below, the section highlighted in green? This is what we will focus on for finding the correct recursive formula!

The question we are working with actually gives us a sequence and we need to find the recursive formula that works with it! Our goal with this question is to work backward to test out each recursive formula given to us until we get the correct sequence. To begin, let’s first identify each term in our given sequence.

As we go through each answer choice, we are looking for the recursive formula that gives us the above sequence 3, 7, 15, 31. Let’s start with choice (1) which happens to be a type of geometric sequence.

Right away we can see the sequence forming for choice (1) is 3, 11, … where the first term, 3, matches our original sequence, but the second term we get which is, 11, does not. This means we will need to move on and find the sequence of the next option, choice (2), which happens to be another type of geometric sequence. Let’s take a look:

For choice (2), we can see that the sequence we get is 3, 7, 127 which matches our given sequence for the first two terms 3, 7, but does not match the third term 127, when we need a 15 here (the original sequence is 3,7,15,31). Thus, we must move onward to the next recursive formula by testing out choice (3), which this time is an arithmetic sequence! Maybe we will have better luck!

This last option, choice (3) provided us with the same sequence we were originally provided within the original question, 3, 7, 15, 31. We have found our answer and now we can celebrate!

## Recursive Formula Examples

Still have questions about recursive formulas? Check out more on recursive formula examples here and in the video above! And if you’re looking to learn all there is to know about sequences, check out this post here! Looking to move ahead? Check out the infinite geometric series lesson here! Also, please don’t hesitate to comment with any questions. Happy calculating! 🙂

## Solving Log Equations: Algebra 2/Trig.

Ahoy math friends! In this post, we are going to focus on solving log equations by solving this Regents questions step by step. We’ll answer this question right away! But if you need more of a review, keep reading and you will find what logarithms are, the different kinds of logarithms rules, and some simpler examples. Ready for our first example?! Check it out below:

## How do I Answer this Question?

Step 1: Let’s re-write the equation to get rid of the “log.”

Step 2: Solve for x in our new equation (5x-1)(1/3)=4

If the above answer makes sense to you, great! If not, that’s ok too, keep reading for a review on solving log equations.

## What are Logarithms?

Logarithms are inverses of exponential equations. Take a look below for a clearer picture.

Basic Formula:

Logarithms can be re-written to get rid of the word “log.” This makes them easier to solve and understand.

Log Rules:

There are a few rules you have to memorize get used to with practice. These rules are used when solving for x in different kinds of algebraic log problems:

Still got questions?  No problem! Check out the videos below and the post here for more on logarithms! Also don’t forget to subscribe below to get the latest FREE math videos, lessons, and practice questions from MathSux. Thanks for stopping by and happy calculating! 🙂

Looking for another math challenge? Check out this post on solving 3 simultaneous equations at once!