October 31, 2006
Today was a pretty easy class. We did a pretest, which wasn't included for marks. Everyone for the most part understood the first page. So I'll just briefly go over those questions first. =)
1. The sum and product of the roots of the quadratic equation 2x2 + 2√(3)x - √(6) = 0 are:
A) √(3) and √(6)
B) -√(3) and √(6)/(2)
C) -2√(3) and -√(6)
D) -√(3) and -√(6)/(2)
The answer would be D. If we remember from previous class lessons, the sum is the negative of b, and the product is the same as c.
2. The Equation x2 + 3x - 2 = 2 has:
A) 2 different negative numbers in its solution set.
B) only 2 solutions and they are both positive.
C) only two solutions
D) no solution
E) none of the above.
The answer for this question was A. This is how we solved it.
x2 + 3x - 2 = 2
x2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4, x = 1
x2 + 3x - 2 = -2
x2 + 3x = 0
x(x+3) = 0
x = 0, x = -3
With these solutions, every single one will work when placed into the equation. Because two of the 4 solutions are negative (-), option A, is therefore true.
3. Solve: (2x)/(x-1) = (x)/(x+1) + 1
For this question, we're solving for x. So basically, you multiple each side by the LCD (Lowest Common Denominator). Which happens to be: 1(x-1)(x+1). By doing that we end up with this.
2x(x+1) = x(x-1) + (x+1)(x-1)
From here on, we solve for X.
2x2 + 2x = x2 - x + x2 - 1
3x = -1
x = -1/3
Therefore the answer to this question is x = -1/3
4. Given the following quadratic equations, use the discriminant to determine which of the following have two real solutions.
A) 3x2 + x - 2 = 0
B) x2 - 2x + 1 = 0
C) -x2 - 12x - 5 = 0
0 <-- ONE REAL SOLUTION.
First we start off with finding the total time it took for the person to get this distance.
Time: 3 h 20 min = 3 1/3 hrs or 10/3 hrs.
Secondly, we need to find AB, and BC.
- We know that by using the pythagorean theorem that a2 + b2 = c2
- Therefore x2 + 52 = AB
- And √(x2 + 25) = AB
- we know the total distance from the point directly across the river from A and C is 10 miles.
- Therefore BC = (10 - x)
Know that we know what AB and BC are, we can look for the equation.
swim time + run time = total time
√(x2 + 25) / 3 + (10 - x) / 5 = 10 / 3
Now we find the LCD again and solve for x!
- First we multiply everything by 15, and we end up with this:
5√(x2 + 25) + 30 - 3x = 50
5√(x2 + 25) = 3x + 20
- To get rid of the radical we have to square both sides!
(5√(x2 + 25))(5√(x2 + 25)) = (3x + 20)(3x + 20)
25(x2 + 25) = 9x2 + 120x + 400
25x2 + 625 = 9x2 + 120x + 400
16x2 - 120x + 225 = 0
- Now that we've found that and it isn't factorable, we use the quadratic formula to solve for x.
x = 120 ± √(14400 - 4(16)(225)) / 2(16)
x = 120 ± √(14400 - 14400) / 32
x = 120 ± √(0) / 32
x = 120 / 32
x = 60 / 16
x = 30 / 8
x = 15 / 4
And there you have it! .. This question covers most of the things we did in this unit. It's challenging, but it won't be on our tests nor our exam. So that's it! ..
Homework was posted by Mr. K earlier.
And our next scribe will be....
CRYSTA! . haha. have fun =)
Remember ... learning is a conversation ... if you're not talking to somebody about it, you're not learning it ... let the conversation begin! ;-)
Have a good Halloween guys! And good luck on your test on wednesday! (:
October 30, 2006
Well time again for yet another BOB..
Well I think for me, and probably most of us, this past unit of Algebra/Radical Equations has flew by. It was a short and straight foward topic. I don't recall having a lot of troubles with this unit because of the simple grasp of solving for x. It actually has been one of my favoured and prefered units. Although, I expect it to get harder and more complex from here. The actual minor part that I may have troubles with is having to check the solutions and not understanding why its an Estraneous Solution. Well, actually it may have been that question our class had stumbled on today (which Mr.k had troubles understanding himself). Lets just hope theres none of those on the test.
Well good luck and good night! :)
also.. Happy advance Halloween!!
I see now that algebra is just an in-depth version of all the equations and formulas we've done before. But I'm wondering why some teacher say that we've been doing algebra since we first started math when my fellow classmates ask when we're starting algebra. I have some difficulty on the absolute value questions that have 2 absolute values on one side of an equation like: |x+1|-|2x-1|=5 (It's a random question I thought of. If it works out properly, cool).
I'm wondering how many cases will there be on a question like that?
That's my BOB for the Algebra Test. Good Night
October 27, 2006
In the last few days we've been looking at radical equations, and when solving these equations (roots) we sometimes get extraneous roots. Well, this class was about: Why do we get these roots in rational equation.
Graphically this is what is going on
In our case here, lets first see the sign changes in the intervals between -infinite to -2; -2 to 1; 1 to 4; and 4 to infinite, in order for us to determine when the equation < style="text-align: center;">
1st interval -ve infinite to -2
pick any number in the interval e.g -12
plug into the formula
October 26, 2006
in class today, for all you naughty kids that skipped (ha, no I'm just kidding), we worked on some problems that Mr. K had put up on the board. We mainly focused on algebraic equations that deal with extraneous solutions and we also had breifly discussed absolute. The four questions that the class worked on were:
x= √x-2 +4
x=6 or x=3
*REMINDER: Always begin solving your algebraic equation by isolating the variable!
To check which answer is correct, we substitute the two solutions we got.
6= √6-2 +4
6= √4 +4
Therefore 6 is the correct answer.
3= √3-2 +4
3= √1 +4
The number 3 does not work out so we reject it. It is an extraneous solution.
√4x+5 - √2x-1= 2
4x+5= 4+4√2x-1 +2x-1
x=5 or x=1
√4(5)+5 - √2(5)-1= 2
√25 - √9= 2
5 -3= 2
Therefore 2 work out and is the correct solution and 1 is the extraneous solution.
5r/ r²-9= 2
(multiply both sides by r-9 to get ride of it on the left)
x= 9/2 or x=-2
Therefore 2 is the correct answer and 9/2 is the extraneous solution.
x cannot equal 0, -4 because you cannot divide by 0!
1/x + 3/x+4= 2(x(x+4))
1(x+4)/x+4(x) + 3(x)/x+4 (x)= 2(x(x+4))
x+4 +3x= 2x²+8x
x= -b ± √b² - 4ac/ 2(b)
x= -2 ± √4-4(1)(-2)/2(2)
x= -2 ± √12/2
x= -2 ± 2√3/2
x= -1 ± √3
*REMINDER: state restrictions before doing anything when dealing with fractions with vaiables in the numerator.
We'll go into more depth tomorrow about absolute numbers, but what we breifly discussed today is:
case 1: x+1= -2
case 2: x+1= 2
|-3 +1| =2
|1 + 1|= 2
Both solutions are possible.
This is not applicable (N.A) because any absolute number must produce a solution that is positive.
case 1: x-4= 2x+1
case 2: x-4= -(2x+1)
5 does not work out, therefore it is an extraneous solution.
Well that's about all we did in class. I hope this helps. Today we were not assigned any excercise, however, our homework for tonight is already posted up on the blog. Last, but not least, the next scribe......hmmmm who will it be next? I say, the SHEEP of the class is, THAT MEANS YOU NATNAEL!!!! muhahaha (evil laugh) take that sheep. I feel a hard class coming up tomorrow!
October 25, 2006
Its me m@rk again. We start off today's short class with four questions.
The first question was:
√x+1 +2=4 REMEMBER: The first step is to isolate the radical by itself
√x+1=2 Square both sides to get rid of the radical sign
x+1=4 Subtract 1 on both sides
x=3 REMEMBER: Substitute your answer back at the equation and see if it makes sense.
The second question was:
x=√x+10 +2 Isolate the radical by itself
x-2=√x+10 Square both sides to get rid of the radical sign
(x-2) (x-2)= x+10 Distribute
x^2-4x+4 =x+10 Simplify it to an equation you are comfortable of doing
x^2-5x-6 = 0 Factor to find the roots
(x-6) (x+1) = 0
∴ x= 6,-1
Remember to substitute your answer in the original equation. In this question the
answer of -1 doesnt make sense, because it will not balance the equation. This case is called an extraneous solution.
The third question was:
4/x=3 , X ≠ 0 REMEMBER: WRITE THE RESTRICTIONS
4/x=3 Multiply both sides by the LCD to get rid of the fraction
4/3=x Look at your resrictions if its not the same then it is not an extraneous solution
Lastly, the fourth question was:
4/x-3/x+1=1 , x ≠ 0,-1 REMEMBER: State your restrictions
4/x-3/x+1=1 Multiply both sides by the LCD
4x+4-3x =x^2+x Simplify
∴ x=2,-2 Look back at your restrictions if its not the same, it is not an extraneous solution.
That's it for my scribe for today. Mr. K , we used the rest of the class time to do our homework. The lady was quite impress on how we focused till the last second of the class.
Our homework for tonight is Exercise #19. The next scribe is......... (background music:dun..dun...) the one and only SAMUS or should I say Mz. Anonymous. Good nyt and have fun.....
October 24, 2006
Today in class Mr.K got us to do some questions on Applications with the Discriminant. Here are the questions and I'll go over them after:
1)For what values of k will the equation: 2x²+4x+(2-k-k²)=0 have exactly one root?
2)3x²-mx+3=0 a) For what value(s) of m will one root be double the other
b) For what value(s) of m will the roots not be real
1) Okay for question #1 we'll use the discriminant formula which is b²-4ac and in order to find k we need to know what the discriminant has to equal to ensure a 1 root parabola and the number is...0. So sub in a, b, and c into the Discriminant Formula
(4)²-4(2)(2-k-k²)=0......................And balance equation
So after all that work we now know that k must equal to 0 in order to have the quadratic equation have only 1 root.
2)First we'll answer b) because it will be used to find a)
b)First we'll plug in the a,b, and c into the Discriminant Formula. We' ll set the equation to zero and solve for m. I'll show you:
m²-36=0..............Notice that m²-36 is a difference of squares?
Now let's look at the values of m on the number line:
Explanation: If we pick a number less than -6 and more than 6 and sub it in for we'll get a positive result but we need a negative number in order to get no roots so the numbers we need are anything between -6 and 6. m=(-6,6)= No roots
a) Okay now for question a. What are the roots in the quadratic formula? They are:
So the equation to find the roots for the questions in 2x#1(the lowest root)=x#2. Now with that in mind we sub in x#1 and x#2 with that equations above and sub in a,b, and c from the original equation and now we have.
Now just solve for m and viola those are the roots.
Okay that's my third scribe and I hope this helps with some questions on Applications with the Discriminant. Homework for Mr.K's class is Exercise #18. The next Scribe will be...m@rk. Good night everybody.
October 23, 2006
We first started with Mr. K giving us three equations.
f(x) = x² + 5x + 6
y= x² -4x - 5
f(x) = x² + 2x - 15
We were to find:
- The ROOTS
- The sum of the roots
- The ROOT PRODUCT
Here were the answers.
f(x) = x² + 5x + 6
ROOTS: x = -3, x = -2
ROOT SUM: (-2) + (-3) = -5
ROOT PRODUCT: (-2)(-3) = 6
y= x² -4x - 5
ROOTS: x = 5, x = -1
ROOT SUM: 5 + (-1) = 4
ROOT PRODUCT: 5(-1) = -5
f(x) = x² + 2x - 15
ROOTS: x = -5, x = 3
ROOT SUM: (-5) + 3 = -2
ROOT PRODUCT: (-5) 3 = -15
Afterwards, we were to compare our answers to the equation. Remember, mathematics is the science of Pattern--so look for a pattern.
We found out that the pattern here was that the root sum is the negative of B. Let's look at the first one as an example.
f(x) = x²
ROOT SUM: (-2) + (-3) = -5
AND the other pattern was that The ROOT PRODUCT is = to C.
f(x) = x² + 5x + 6
ROOT PRODUCT: (-2)(-3) = 6
Knowing that there's a relation between them, we can easily figure out the roots of an equation just by simply looking at it. There's a catch though..
Mr. K then gave us another equation and asked us to find the roots of it, the sum of the roots and the product of the roots.
f(x) = x² + 6x + 10
This equation can't be factored out perfectly. it doesn't have REAL roots so it must have an imaginary root. But, since it's not factorable, the other thing we can do is use the Quadratic Formula.
REMINDER: i² = -1 OR i = √-1
x1 = -3x + i
x2 = -3x - i
^those are your roots. To find out the product of the roots you must distribute -3x - i & -3x + i.
( -3x + i )( -3 - i )
9 + 3i - 3i - i²
9 - i²
9 - (-1) = 10 <- The product of roots.
It's different when there is a number in front of a.
y(x) = 2x² + 5x - 3 <- You can't just take the root sum or root product just by looking at the equation at this point. Since there's a number in front of a it'll factor the other numbers resulting in a different number..
So to do this correctly you must find the coefficients and rewrite 5x. In this case it's going to be 2 & 3.
-1 + (-3)/2
-2/2 + (-3)/2
We should be also aware of number relationships. Here's a diagram for refrence.
In a quadratic formula, if the inside of a radical is negative then there are no roots. it discriminates the qualities of the equation.
b² - 4ac > 0 = ROOTS
b² - 4ac = 0 = 1 ROOT
b² - 4ac < 0 =" NO">
HOMEWORK IS EXERCISE 16
NEXT SCRIBE WILL BE.................benofschool
October 20, 2006
Mr. K was away today so we had a sub. We only worked on Exercise 17 for the whole class.
I don't know what else to put down because that's all we did. No major problems or anything with the work, I believe.
Mr. K, if your reading, just to let you know, our class was very well behaved and we did our work nice and quietly. Too quiet actually.
I guess that's all for my scribe..I don't feel this is fair, so I'll just scribe again on Monday.
Have a good weekend guys.
October 19, 2006
Today we continued from yesterdays lesson and we started with this question.
1. a2 - 3a - 15 = 0
To solve this, we had to use the quadratic formula.
a = 3 ± √(9 + -4(1)(-15))
a = 3 ± √(69)
(a - 5)(a = 3) = 0
a = 5, a = -3
^^ Here, instead of using the quadratic formula (which would've worked as well) we simply factored it. Like Mr. K says, why use the sledge hammer when we can use the hammer?
Lets stop for a moment and see if we recognize any patterns in the following equation.
3. (x - 2/x)2 - 2(x - 2/x) - 15 = 0
This is something we call a quadratic in form.
If we're to do something like this:
Let a = x - 2/x
we end up with an equation that looks like this:
a2 - 2a - 15 = 0
Which happens to be the exact same equation as the one we did above. So have you seen the pattern??
Above, we figured out that the roots of the equation a2 - 2a - 15 = 0 were a = 5 and a = -3.
We Let a = x - 2/x as well.
So lets plug in what we need to and solve for question 3.
a = 5
x - 2/x = 5
Now multiply both sides by x to get rid of the x in 2/x.
x2 - 2 = 5x
x2 - 5x - 2 = 0
^^ Since that doesn't factor nicely, this is where we can use the quadratic formula.
x = -b ± √(b2 - 4ac
x = 5 ± √(25 - 4(1)(-2))
x = 5 ± √(23)
x1 = 5 + √(23)
x2 = 5 - √(23)
a = -3
For this one, we do it exactly the same way.
If you guys want to try it out and want to see if you did it right, here's the answer.
x = 3 ± √(17)
4. x4 - 132 + 36 = 0
- Let a = x2
- a2 - 13a + 36 = 0
- (a - 9)(a = 4)
- a = 9, a = -4
It should be pretty straight forward on how to get there, but if you have any questions, feel free to ask! =)
Anyway, this is where most people do this:
x2 = 9, x2 = 4
x = 3, -3, x = 2, -2
9 OUT OF 10 TIMES when you solve it this way, you'll forget the negative (-) solution
Lets get into the habit of solving our equation this way:
x2 - 9 = 0, x2 - 4 = 0
(x + 3)(x - 3) = 0, (x + 2)(x - 2) = 0
x = -3 and 3 , x = -2 and 2
This way, there's no way you'll forget the negative (-) solution.
Okay. I hope this helped Sam! =) Get well soon! .. If you have any questions don't be shy to ask. :D
NEXT SCRIBE IS .. How'd you guess? MERIAN.
Homework tonight is Exercise 15. Have fun =)
October 18, 2006
Here it is explained.
You can practice using it over there!
Now sing it with me folks!
x equals negative b
plus or minus the square root
of b squared minus 4ac
all over 2a
FIND THE ROOTS USING 'COMPLETING THE SQUARE':
* * Okay so first of all, we simply transfored our original equation from general form to standard form.
- NOW WE MUST ALLOW Y = 0
Is there an EASIER APPROACH to find roots of all parabolas of the general form?
- YES. Yes there is!
Quick Note: y=ax²+bx+c
- represents ALL, yes every single parabola out there
CONVERTING THE GENERAL FORM TO THE STANDARD FORM FOR THE EQUATION OF A PARABOLA:
i) The General Form
ii) Factor "a" out of all x-terms
iii) Find half the coefficient of the x-term & square it
1/2(b/a)=b/2a (b/2a)² = b²/4a²
iv) Complete the square
v) Factor and Simplify
vi) Distribute "a". Find common denominator
vii) Add Fractions
**YEAH, THATS UGLY
FINDING THE ROOTS OF A PARABOLA FROM THE EQUATION IN GENERAL FORM CONTINUED FROM ABOVE:
i) Set to "zero"
0=a(x+b/2a)² + -b²+4ac/4a
ii) Add b²+4ac/4a to both sides
iii) Mulitiply both sides by 1/a
iv) Find the square root of both sides
[b²-4ac/4a²]^1/2 = x+b/2a
v) Simplify Radical
vi) Add -b/2a to both sides
vii) We add the fractions and get
(dun dun dun..) THE QUADRATIC FORMULA!
REMEMINDER: It is important to remember this specific formula!
So for the finale of my scribe- heres our handy sing along.
(8).. X equals negative b,
plus or minus the square root of b square,
minus 4ac, all over 2a
REMINDER: In order to transfer information from short term memory into long term memory, take in the information SIX TIMES.
- Homework assignment: Exercise #14
- - Nightt :)
October 14, 2006
* Represents a beginning to a new page.
Examples are in purple.
Using the Factored Form:
1. You can quickly find the roots by solving the quadratic.
example: 0= x² + 5x + 6
x= -2 x= -3
Roots at x= -2, x= -3
2. You can find the x-coordinate of the vertex by finding the midpoint of the roots.
example: Xv= (-2) + (-3) / 2
3. Once you know the x-coordinate of the vertex you can substitute back into the equation to find the y-coordinate of the vertex.
example: F(-5/2) = (-5/2)² + 5(-5/2) + 6
F(-5/2) = -1/4
Therefore the vertex is at (-5/21, -1/4)
The General Form for the Equation of a Parabola:
F(x) = ax² + bx + c
The Role of Parameter b:
b produces an obligue translation.
The Role of Parameter c:
c is the y-intercept of the parabola. It's coordinates are (0,0)
*Completing the Square:
1. Factor the coefficents of x² out of the two
terms. Leave some "space" in the brackets.
y= 3x² + 12x -5
y= 3(x² + 4x )-5
2. Divide the coefficent of the x term by 2 and square the result.
4/2 = 2
2² = 4
3. Add and subtract the result from step2 inside
the brackets [you just "completed the square"]
y= 3(x² + 4x + 4x -4x) -5
4. Factor the first three terms in the brackets into
a perfect square.
y= 3[(x+2)² -4] -5
5. Distribute the number outside the brackets.
y=3(x+2)² -12 -5
F(x) = AsinB(x-C) + D or F(x)= ACos B(x-C) + D
The Basic Function Graphs:
The Role of Parameter A:
/A/ is called the Amplitude or the "vertical strecth." It tells you the distance from the sinusoidal axis to the maximum or minimum value.
The Role of Paramere B: