Word Problems With Quadratics

Thursday

On Thursday, 4/4, we learnt about solving word problems that would require us to use the quadratic formula. To refresh your memory, if it's not already memorized, here is the quadratic formula:

At the beginning of class we split off into groups that we would later make notability projects with. Before that we did the homework problems up on the board.

This is the problem that my group worked on:

A deck of uniform width has area 72m^2 and surrounds a swimming pool that is 8m wide and 10m long. Find the width of the deck.

It always help to draw out the problem to figure out how to solve it.

Above are the steps we went through to solve the problem using the equation "length x width = area". Our length is 8 + 2x because we know that the pool is 8 m wide but there are also two areas of x on both side of the pool. The width is 10 + 2x for the same reason. When we multiply these two we get 72 + 80 because the problem says the deck's area is 72 and we can find the area of the pool by multiplying 8 and 10. After simplifying the problem we get an equation that we plug into the quadratic equation and simplify until we get x = (-9 +/- 3sqrt17)/2, since lengths can be negative, we know that the sign before the root is +, not -. We plugged this into our calculator and got close to 5/3.

After doing these problems we sat down with our groups and started working on more word problems.

Friday

On Friday we finished on our group word problems. Here is the problem that my group worked on, shown in a notability note:

This problem was solved much like the example I showed before, we set up the problem much like any other word problem, trying to solve the x variable, and then when we got to the quadratic we plugged it into the quadratic formula and simplified some more. Since this is the blog you can't hear the recording that is in our project but if you want to hear it, or see any of the other projects from other groups I would recommend going and checking out the dropbox where the all are.

Thanks for reading, hope you learnt a lot!

-Rowan

The Quadratic Formula and Imaginary Numbers

In the past two days of class, we touched on two major topics:

-How to apply the Quadratic Formula to quadratic equations, along with the parabolas that accompany them (along with the memorization of the formula);
-Imaginary numbers, what they are, what they mean and how to apply them in mathematics.

So, to dive right in--

The Quadratic Formula
As I'm sure you have heard before, the Quadratic Formula is a formula that allows you to plug in values "a", "b", and "c" from an equation, solve it, and from there obtain your x values.

Here it is:

This equation is used to find the value of x in a quadratic example.

a=coefficient of x^2
b=coefficient of middle term x
c=third term

You simply plug these values in for a particular equation, solve it, and out comes your (usually) two x values.

An Example:

The first step of this is to determine the value of y. Say y is zero, and the rest is easy. Here are the values for a, b, and c:

a=1

b=-10

c=9

Then, you plug these values into the quadratic equation (above). This comes out to:

x=(10 +/- sqrt[100-36])/2

x= (10+/- 8)/2

x= 9, 1

I hope this quick overview of the Quadratic Formula has helped in solidifying your understanding. 

Quadratics in Parabolas

Every quadratic equation, when graphed, comes out to a parabola; meaning that it resembles some kind of a "half-pipe" in its line. There are three kinds of parabolas in quadratics (disregarding sideways vs vertical), displayed below:

Let's take a look at the graph of the quadratic solved previously in the post;

As you can see, the quadratic x^2-10x+9 is a parabola as it resembles a half-pipe. If you take a look at where the line intersects the x-axis, you'll see that those points are the two solutions of the equation. This applies to all quadratics looking for an x value.

The number of solutions a quadratic has (one, two, or zero) is based on the discriminant in the Quadratic Formula. This is the b^2-4ac, of course with the a, b, and c values plugged in the with the relative equation.

Here is a chart developed in class that distinguishes between one, two, or zero solutions in a quadratic:

Note that the "one x-intercept (solution)" side includes two intercepts as well. No x-intercepts include negative square roots.

Imaginary Numbers
Contrary to its name (and the several puns that accompany it; I'll spare them here, for the most part) imaginary numbers are very much real.

They are represented by i; an abbreviation for a definition that is actually very simple:

i=sqrt(-1).

Yup, that's it.

Here is a list that we brainstormed in class about imaginary numbers and what they mean (after a quick internet source on our fancy imaginary--I mean i--Pads).

Here is a diagram with the explanation of what "complex numbers" really are (we didn't seem to understand it at first):

And some quick work with i and its exponents, following a remarkable pattern:
i=i
i^2=-1
i^3=-i
i^4=1
i^5=i

I hope this post has helped your understanding of Quadratics, Parabolas, and our newly-discovered topic of imaginary numbers.

--Peter

Solving Radical Equations

Hello Class,
Sorry for the late post but here is all the information for March 27th 2013 and March 28th.  I will show you some step by step examples:


Check out this video for more information on Sums of Radicals
http://m.youtube.com/watch?v=ZA497RyAKe8

Here is the homework or practice problems that you can try out :)

Page 259: problems, 1-25 (every other odd)

And page 261: problems, 1-25 (every other odd)

Hre is the class information to Solving Radical
 equations. Here is a video to start you out. :)
http://m.youtube.com/watch?v=LY8VBsLf-4M

Here is the homework or practice problems that you can try out :)

Page 261: problems, 1-25 (every other odd)

Page 259: problems, 1-25 (every other odd)

Solving Equations with Roots

On Thursday we were finished with our self-guided projects on 5-9 and 5-10 and moved on the more math. Thursday also happened to be Pi day so we had some very good cookies and had some Pi related fun. We practiced reciting the digits from Pi, everyone did fairly well, and two students reached of 40 digits!

When we were done with our fun we moved on to learning about solving equations that had roots in them.

This little symbol is the square root sign. The solution then becomes whatever will square to become x. For example, the square root of 36 is 6, because 6^2 = 36.

Here is another image which displays the “index” part of roots. Whatever is in the index is the exponent that the solution would have to equal the radicand. In the example above, the solution is 3, because 3^3 = 27.

Here is an example of a problem where we simplified the square root of 32. 32 is not a perfect square, so we had to find a perfect square that was a factor of 32. 2 and 16 multiply to 32 so we can separate the two and then find the square root of 16, which is 4. So that means our simplified expression is 4*sqrt2.

In these four problems all you need to do is simplify within the root. In the third problem, we see -1^2 become 1, this is because when two negatives multiply, they become positive. When you have one term by itself that is squared, you can cancel it, but not if there are multiple terms. In problem one you can’t cancel because there is subtraction but in problem 4 you can cancel the square.

When you have fractions inside a root symbol, you can simplify the numerator and the denominator. We can see in problem 2 an example of solving with root 3. 3*3 equals 9, and 9*3 equals 27 so 27 becomes 3. 5*5 equals 25 and 25*5 equals 125 so 125 becomes 5. The simplified expression is 3/5.

Thanks for reading my blog! I hope you learned a lot!

Adding and Subtracting Radicals

DAY 1   

On Wednesday, March 20, we worked with square roots again. However, instead of multiplying them we added the roots. At the beginning of class we were asked to solve four problems on the board. The first was

We solved the first problem by factoring and then combining common terms. 

The next problem was a bit more difficult because it contained division. Also there were two different methods of solving it.   Here are the two methods:

                                     

In the third problem we had to use the foil method and learned that you can only multiply out numbers with the same index. 

Example of multiplying roots with (1) the same index and (2) different indices.

Example of solving a problem using the foil method when the numbers have different index numbers.

The final problem introduced variables and fractions into the mix. Treat the variables like any other number. To deal with the fractions, you must make all the denominators the same. The easiest way to do this is to find the least common denominator, or the smallest whole number that is divisible by all the denominators in the problem. This is the same idea behind finding the least common multiple for whole numbers. Once you find the least common denominator, multiply the original denominator by the least common denominator divided by that original denominator (New Denominator = LCD). Then multiply the numerator of the preceding fraction by the LCD divided by the denominator (New Numerator = (LCD ÷ Denominator) x Numerator.

 For example

However,  if this question were in a textbook, the preceding answer would not be acceptable because it is not rationalized. Rationalizing is making sure that there is no root in the denominator. To rationalize our answer,  you must multiply the numerator and the denominator by the denominator. For example

After learning all the above, the class put our knowledge to the test by solving the following more challenging problem:

If there is a regular octagon inscribed in a square that has lengths of 4, find the length of one of the octagon's sides.

   

To solve the problem take the following steps:

1. Label as x the legs of the isosceles triangles found in each corner of the square.                                    

2. Since the isosceles triangle are right triangles, label the hypotenuses √2x^2.

3. Given that each side of a square is 4 and the sum of the lengths of the legs of the two isosceles triangles that make up part of that side of the square are equal to 2x, 4-2x would equal a side of the octagon.

4. If the octagon is regular then √2x^2 = 4 - 2x.

5. √2x^2 = x√2

6. x√2 + 2x = 4

7. x(√2 + 2) = 4

8. x = 4/(√2 + 2) = 1.17

9. The length of each side of the octagon is 2.7378.

Although this answer is correct, there is another possible answer. To find this we must use quadratics. The quadratic formula is a way to factor quadratics.   

Quadratic Formula

In the quadratic formula, a is the coefficient of x^2, b is coefficient of x, and c is the number that is not multiplied by a variable. For example

Now that we have reviewed quadratics back to our problem of determining the sides of the octagon.

1. If the octagon is regular then √2x^2 = 4 - 2x.

2. 2x^2 = (4 - 2x)(4 - 2x)

3. 2x^2 = 4x^2 - 16x + 16 

4. 0 = 2x^2 - 16x + 16

5. 0 = x^2 - 8x +8

6. x = (8 - √64 - 4 (1)(8))/2 or  x = (8 = √64 - 4 (1)(8))/2  (using quadratic formula)

7. x = (8 - √32)/2  or x = (8 + √32)/2

8. x = (8 - 4√2)/2 or x = (8 + 4√2)/2  

9. x = 4 - 2√2 or x = 4 + 2√2

10. x = 6.82 or x = 1.17

11. If you substitute 6.82 in the equation 4-2x, the side length is negative, which is impossible. If you substitute 6.82 in the equation √2x^2, the side length is 9.64, which makes the octagon larger than the square it is inscribed in. Therefore, x needs to equal 1.17.

11. The length of each side equals 2.7378.

Here is a video that reviews the basics of adding and subtracting radicals:

Simplifying and Multiplying Rational Expressions

On tuesday we started out the class everyone's favorite way - a surprise homework collection! But once everyone was done stressing about how they did, we moved onto a new topic, simplifying rational expressions. This is similar to simplifying fractions, but with variables involved. You can cancel out any terms that the numerator and denominator have in common, just like one would if you were simplifying even the simplest of fractions, such as 5/20. 5 and 20 are both divisible by five, so if you divide both by five the fraction simplifies to 1/4.

Once we added variables to the mix, it became more complex, because we almost always had to factor the numerator and the denominator in order for the terms to cancel. The first example we used is in the picture below.

As you can see, once you factor both expressions, any factors that they have in common will cancel. It is important to remember, however, that you can't cancel any terms until each term is connected by multiplication. If one of the terms has addition or subtraction in it, such as (x-2), you can only cancel the full (x-2) if both expressions have it in common. You can't cancel just the x or just the -2.

After everyone understood the idea of simplifying rational expressions (with variables), we moved on to also finding the domain and zeros of the simplified expression. We had worked with domain before, when we were talking about graphing earlier in the year, but just as a review we defined it again.

Domain: The set of all x values that work for the expression (in order to have a valid output).
An example of an invalid output would be any x values that would make the denominator 0, because then the expression would be "undefined".

We have not worked with zeros before, so this was a new definition for us.

Zeros: All x values that make a fraction equal 0.
This means any x value that will make the numerator 0, because any fraction with a numerator of 0 is equal to 0.

In this example, the two expressions simplified to (x+1)/(x-1). This means that the number that would make the denominator 0 is 1, which is why it is excluded from the domain. All other numbers work with x-1. The zero, or the number that makes the numerator 0, was -1, because the numerator was x+1, and if you add 1 to -1, the difference is 0.

We spent the rest of the class working on examples and finding the domain and zeros of different rational expressions. This was the gateway into Wednesday's class.


We started the class on Wednesday by going over the homework, which was all either simplifying fractions with variables or finding the domain and zero. One that people had a particular amount of trouble with was #31. Mainly the problem was factoring each of the expressions, because once they were factored the problem was actually fairly simple.

The denominator of the fraction could be factored using the difference of squares pattern that we learned earlier: a^2-b^2=(a-b)(a+b). Eventually, part of the numerator could be factored using the same pattern. Once everything was completely factored, everything in the denominator factored out, and only x+y remained in the numerator. That meant that the simplified answer was (x+y)/1, which is equal to just x+y.

On Wednesday, we expanded on the idea of simplifying these expressions by adding multiplication and division to the mix. In order to refresh our memory, Lisa gave us an example of a multiplication problem with two fractions without any variables. She showed us that you could simplify the fraction after you multiply, or you could simplify the fractions before multiplying in order to save yourself time and work.

Dividing fractions is very similar to multiplying fractions, because you can turn those problems into multiplication problems. The process is KCF, or Keep Change Flip. You keep the first fraction in your equation, change your sign from division to multiplication, and then flip your second fraction.

Since we all understood the basic principles of multiplying and dividing fractions, we quickly moved on to adding variables into the fractions. The process is similar, but you just need to factor the numerators and denominators of each fraction. That way you can cancel the terms before you do the actual multiplication, like in the second example in the picture above. A term that is in the numerator and the denominator of the same fraction will cancel, or a term that is in the numerator of one fraction and the denominator of another. A term that is on the same side of the fractions will never cancel. Once everything is cancelled, you can multiply to find your answer. I found a video on Youtube that explains multiplication of "algebraic fractions."

 Hope this clarified any questions you had!

- Molly Brock

Polynomial Equations for Problem Solving and Review

On thursday, our class schedule was organized to help us review the previous concept we had learned, quadratics, and learn a final concept before it would be time to review for our upcoming test.

Some of the problems in the homework that many students had difficulty with were the more complex problems, such as problem number 41, in which we had to figure out what possible numbers could be represented by the variable, t.

Lisa explained that to solve this problem, we had to first expand (t+1)^3. An easy way to expand exponents is to apply the problem to pascals triangle. For an exponent of 3, the numbers that the  variables are multiplied by are 1, 3, 3, and 1. The number of times that 1, 3, 3, and 1 are multiplied by the variable decreases by 1 with each set in the series, starting by the largest original exponent, 3.
After expanding (t+1)^3 to t^3+3t^2+3t+1, we found that t^3 could be eliminated from both sides of the equation, as the other side of the equation was t^3 + 7. We also subtracted 7 from the right side of the equation so it could be left with 0. The equation now read 3t^2+3t-6=0, which was divisible by 3. Dividing the problem by 3 left us with a problem that we could factor to (t+2)(t-1)=0. Therefor, t was either equal to -2 or 1, or {-2,1}


For our previous homework assignment we had also been introduced a word problem. We were unfamiliar with the concept of forming an equation based on measurements in this way, so Lisa further introduced the concept in class, starting by explaining the word problem from our homework:
#5, page 189: Find the dimensions of a rectangular lot if its length is 7 m greater than its width and each of its diagonals is 13 m long.

Lisa showed us that an easy way to approach a problem like this was to draw a picture. We drew the rectangle and labelled the dimensions. The width was labelled x to represent its length and the length was labelled x+7 because we knew that the length was 7 m greater than the width. Next we labelled the diagonals as 13 because we knew that the diagonals were 13 m long. The length, width, and diagonal formed a triangle that we could use to solve the problem. We plugged in the information we had to the equation to find the area of a triangle:

a^2+b^2=c^2   --->   x^2+(x+7)^2=13^2 

We then expanded the problem and subtracted 169 from each side of the equation to bring us to 2x^2+14x-120=0, which was divisible by 2. We were able to factor x^2+7x-60=0 to (x+12)(x-5)=0. Therefor, x was equal to -12 or 5, and because the problem dealt with measurement, x had to be positive. 

So the width being x, was 5, and the length, being x+7, was 12.
Answer: The lot is 5x12 m.

We continued to practice similar word problems to number 5, such as number 7:
Two ships leave port, one sailing due west and the other due south. Some time later they are 17 miles apart and one is 7 miles farther from the port than the other. How far is each from port?

We started by drawing the port as a point and drawing the distance that each ship had travelled. We then plugged in the distance each boat had travelled from the port and the boats' distances from each other to the formula for the area of a triangle. The equation was:

x^2+(x+7)^2=17^2

We then simplified the equation as we had done with the previous one until we figured out that x was equal to 8, so the answer was: one boat is 15 miles and the other is 8 miles from the port. 

On Friday, our class schedule was similar to Thursday's. We first reviewed our homework, which had been review for the test. Lisa presented us with the topics that the test would cover:

We had learned all the topics in previous weeks and had solidified them through practice. We had not, however, fully explored word problems, so we next focused on some more difficult problems than what we had worked on on Thursday, such as problem number 15 on page 190:

When a 0.5 cm was planed off each of the six faces of a wooden cube, its volume was decreased by 169 cm^3. Find its new volume. 

Students had difficulty approaching this problem, so Lisa gave us pieces of paper and told us to physically cut off a square from each of their corners so they could be folded into a box. 

We then drew the box and established that each side had originally been equal to x but was now equal to x-1, because 0.5 cm had been planed off of each side. The area of a cube is length times hight times width, so we plugged in the information we had to establish the volume:

(x-1)^3

We used the exponent three because we knew that each side of a cube is the same.

We also knew that the new volume was equal to the old volume (x^3) minus 169cm^3. We combined the two equations to form a final equation: 

x^3-169=(x-1)^3

We were able to establish our answer having been given this strategy to approach the word problem.

Lisa explained that the test we were about to take would not cover word problems that were this complicated, but it was helpful and interesting to explore them in such a way. We left class prepared to review for our test on Tuesday.

When struggling with a concept we learn in class during homework, I find that the Algebra and Trigonometry book have clear explanations to the subjects we are covering. 

I found that the examples in the book on page 188 laid out all the steps to solving the word problems we were working on in class in a clear way, and I was able to apply it to my own thinking.

Thanks for reading! I hope you found this helpful.

Sadie Grant