The nature of quadratic roots
(b²-4ac) is the discriminant

If b²-4ac is The equation has:

> 0 and a non perfect square - 2 irrational roots

>0 and a perfect square - 2 rational roots

=0 - i real root

<0 - 0 real root, imaginary roots


Example:

Given x² - 6x + 21 = 0

What type of roots does it have?

a= 1
b= -6
c= 21

Plug it in:

X= (-b +- sqrt(b²-4ac))/(2a)

= -(-6) +- sqrt(-6²)-(4)(1)(21)
= 6 +- sqrt(-48)/2
= 6 +- sqrt((-1)(16)(3))/2
= 3 +- 4i sqrt(3)

Therefore this equation is an example of b²-4ac < 0 and has imaginary roots.

Continuation of our last topic …

Given the equation

If the roots of the equation are known, the equation can be found,it is...

x²+ (-sum of roots)x + (product of roots) = 0

x² + -b/a x + c/a = 0

Example 1

GIVEN THE ROOTS :( 1 ½, 7/5) state the quadratic equation

1. Find the sum
r1 +r2 = 1 ½ + 7/5
= 3/2 +7/5
= 15+14/10
= 29/10
-b/a= 29/10
b/a= -29/10

2. Find the product
r1*r2 = (3/2)(7/5)
= 21/10
c/a = 21/10

3. Plug it in
x² + -b/a x + c/a = 0
x² - 29/10 x + 21/10 = 0

Quadratic Equation = 10x² -29x +21 =0

Example 2

GIVEN THE ROOTS :( 1+2i, 1-2i) state the quadratic equation

1. Find the sum
r1 +r2 = (1+2i)+(1-2i)
= 2
-b/a= 2
b/a= -2

2. Find the product
r1*r2 = (1+2i)(1-2i)
= 1 + 2i - 2i + 4i²
= 1 - 4i²
= 1 – 4 (-1)
= 5
c/a = 5

3. Plug it in
x² + -b/a x + c/a = 0

Quadratic Equation = x² -2x + 5 =0

The next scribe is Djannine.

Guys if u have any questions, don’t be shy to ask for my help. Thank you. Have a good day!

quadratic function

hi this is john.. I'm the scribe of the day.. we started our lesson by discussing writing a quadratic equation when roots is given..

WRITING A QUADRATIC EQUATION WHEN ROOTS IS GIVEN

If the roots of the equation are known, the equation can be found,it is...
x²+ (-sum of roots)x + (product of roots) = 0
Finding the sum of roots using Quadratic Formula = -b + Ö b²-4ac + -b - Ö b²-4ac2a 2a
= -b + Ö b -4ac -b-Ö b-4ac2a
= -2b = -b2a a
SUM OF ROOTS = R1 + R2 = -b/a

Finding the product of roots = (-b +Ö b²-4ac / 2a) (-b-Ö b²-4ac / 2a)
=b² - b Ö b²-4ac + b Ö b²-4ac - (b²-4ac)4a²
= b ²- (b²-4ac)4a²= 4ac / 4a²= c /a

PRODUCT OF ROOTS = R1*R2 = c /a

It came from the formula of the quadratic equation
x² + (-sum of roots) x + (product of roots) = 0
x² + -b/a x + c/a = 0

Example: Given the roots(-2,4) state the quadratic equation

1)Find the sum of the roots = -b/aSum=r1+r2 =-2+4=2 -b/a=2 so b/a=-2

2)Find the product=r1*r2 = (-2)(4)= -8 -8=c/a

3)Plug in x² + (-b/a) x + c/a=0x²+(-2)x +-8 =0
x² -2x - 8=0

Ex: Given the two roots (-6,-1) state the quadratic equation

Sum of roots -6 + -1= -7-b/a=-7 so b/a=7

Find the product =r1*r2 = (-6 -1)=6 so c/a=6x² + -b/a x + c/a=0

check:(x+6)(x+1)=0
roots x=-6x=-1

Ex.)GIVEN THE ROOTS:(7+ Ö 13 , -7 Ö 13) state the quadratic equation

SUM= 7+ Ö 13 + 7- Ö 13 =14/3 =-b/a so b/a=-14/33 3

Product =(7+ Ö 13) (7- Ö 13) = 49 -7 Ö 13+ 7 Ö 13 -13 = 36/9 = 4
3 3 9
so c/a = 4

the equation x² + -b/a x + c/a =0 x² -14/3 x + 4=0

becomesQuadratic Equation = 3x² -14x +12 =0

REMEMBER: When Given the two roots of a quadratic,the equation is [x² + -b/a x + c/a =0]where -b/a =the sum of the roots and c/a =the product of the roots.

The next scribe is luis.

Classmates if u have question..dont be shy to ask for my help because im having a hard time doing this blog. thanks.

Quadratic Function Project

This is the project for the first unit. Good job. Enjoy

3 methods to finding roots

Okay, so today we were going over 3 different ways to calculate the roots of an equation. They were: factoring, completing the square and the quadratic formula.

Let's take the example:

5x²+7x+2=0

For the first method we factored the equation(you know how to do this).
You factor out like:
(5x+2)(x+1)=0
To factor an equation with a coefficient in front of the variable you expand it like so:

5x²+7x+2=0

5x²+5x+2x+2=0

5x(x+1)+2(x+1)=0

(5x+2)(x+1)=0

Now that you have that, we calculate the roots. For the first bracket, we isolate the 5x like so:
5x+2=0
5x= -2
Get x by itself to get:
x= -2/5

And this is your first root. The second is easier. Take (x+1) and figure out what number x can be to make it equal 0. That would make the second root, -1.
The two roots of 5x²+7x+2=0 are x= -2/5, -1.

The second method we did was Completing the Square(which we did earlier this week). It requires a little more time than the other methods but if done right, the results will be the same.
Let's take our example again:
5x²+7x+2=0

1. We factor out 5 to get x² on its own:
x²+7/5x+2/5=0

2. Now we focus on x²+7/5 and divide 7/5 by 2 and squaring it.
x²+7/5 --> Dividing by 2 in a fraction is like multiplying the denominator by 2. --> 7/5(1/2)
7/5 --> 7/10 then we square it --> (7/10)² = 49/100

3. We plug that into our equation as --> (x² + 7/5x + 49/100) +2/5 - 49/100=0
and factor it into --> (x+7/10)² +2/5 - 49/100=0

4. Now, let's transpose the last two terms so the equation is like this:
(x+7/10)² +2/5 - 49/100 --> (x+7/10)² = 49/100 - 2/5
Find the LCD between 49/100 and - 2/5 --> 49/100 - 40/100

5. Now you will have: (x+7/10)² = 9/100 so now you can square both sides:
√(x+7/10)² = + or-√9/100 (since it is a squared number, you can’t be sure if it is positive or negative)
x+7/10= + or - 3/10

6. Now you can calculate the roots--> transpose 7/10
--> x= -7/1o + 3/10 = -4/10= -2/5 <-- That is the first root.
x= -7/10 -3/10 = -10/10 = -1

The roots are x= -2/5, -1.

And lastly, Method No.3, the Quadratic formula which was also learned this week.
Once again, take 5x²+7x+2 and use the quadratic formula which is:
x= (-b +/-√b²-4ac) ÷ 2a

a= 5, b= 7, c= 2 Now plug that into the equation.

x= (-7 +/- (√7² - 4(5)(2)) ÷ 2(5) -->
x= -7 +/- (√49- 40) ÷ 10 -->
x= -7 +/- (√9) ÷ 10 -->
x= -7 +/- 3 ÷ 10 --> If it helps, you can look at it like: -7/10 +/- 3/10
Find the first root --> x= -7/10 + 3/10 = -4/10 = -2/5
The second root now --> x= -7/10 - 3/10 = -10/10 = -1

So after using all three methods, the roots will be x= -2/5, -1.

And thus is my blog for the day. If you have any problems or find my explanations a little hard to understand, feel free to make any comments. It was my first time so it might be a bit unorganized.

And now the next scribe will be… John…

Okay, so today we were going over 3 different ways to calculate the roots of an equation. They were: factoring, completing the square and the quadratic formula.

Let's take the example:

5x²+7x+2=0

For the first method we factored the equation(you know how to do this).
You factor out like:
(5x+2)(x+1)=0
To factor an equation with a coefficient in front of the variable you expand it like so:

5x²+7x+2=0

5x²+5x+2x+2=0

5x(x+1)+2(x+1)=0

(5x+2)(x+1)=0

Now that you have that, we calculate the roots. For the first bracket, we isolate the 5x like so:
5x+2=0
5x= -2
Get x by itself to get:
x= -2/5

And this is your first root. The second is easier. Take (x+1) and figure out what number x can be to make it equal 0. That would make the second root, -1.
The two roots of 5x²+7x+2=0 are x= -2/5, -1.

The second method we did was Completing the Square(which we did earlier this week). It requires a little more time than the other methods but if done right, the results will be the same.
Let's take our example again:
5x²+7x+2=0

1. We factor out 5 to get x² on its own:
x²+7/5x+2/5=0

2. Now we focus on x²+7/5 and divide 7/5 by 2 and squaring it.
x²+7/5 --> Dividing by 2 in a fraction is like multiplying the denominator by 2. --> 7/5(1/2)
7/5 --> 7/10 then we square it --> (7/10)² = 49/100

3. We plug that into our equation as --> (x² + 7/5x + 49/100) +2/5 - 49/100=0
and factor it into --> (x+7/10)² +2/5 - 49/100=0

4. Now, let's transpose the last two terms so the equation is like this:
(x+7/10)² +2/5 - 49/100 --> (x+7/10)² = 49/100 - 2/5
Find the LCD between 49/100 and - 2/5 --> 49/100 - 40/100

5. Now you will have: (x+7/10)² = 9/100 so now you can square both sides:
√(x+7/10)² = + or-√9/100 (since it is a squared number, you can’t be sure if it is positive or negative)
x+7/10= + or - 3/10

6. Now you can calculate the roots--> transpose 7/10
--> x= -7/1o + 3/10 = -4/10= -2/5 <-- That is the first root.
x= -7/10 -3/10 = -10/10 = -1

The roots are x= -2/5, -1.

And lastly, Method No.3, the Quadratic formula which was also learned this week.
Once again, take 5x²+7x+2 and use the quadratic formula which is:
x= (-b +/-√b²-4ac) ÷ 2a

a= 5, b= 7, c= 2 Now plug that into the equation.

x= (-7 +/- (√7² - 4(5)(2)) ÷ 2(5) -->
x= -7 +/- (√49- 40) ÷ 10 -->
x= -7 +/- (√9) ÷ 10 -->
x= -7 +/- 3 ÷ 10 --> If it helps, you can look at it like: -7/10 +/- 3/10
Find the first root --> x= -7/10 + 3/10 = -4/10 = -2/5
The second root now --> x= -7/10 - 3/10 = -10/10 = -1

So after using all three methods, the roots will be x= -2/5, -1.

And thus is my blog for the day. If you have any problems or find my explanations a little hard to understand, feel free to make any comments. It was my first time so it might be a bit unorganized.

And now the next scribe will be… John…

Axis of Symmetry of the Quadratic Formula

Yesterday we learned about the Quadratic Formula, which lets us turn ax²+bx+c=0, into: x=(-b÷2a)+/-(√b²-4ac)÷2a. We learned how to take the roots(where the line crosses the x axis) from that equation.

Now today, we learned how to take the Axis of Symmetry(the line which seperates the graph to create a mirror image) from that equation also.

The Axis of Symmetry (AoS from now on) is taken from the equation directly.

x=(-b÷2a)+/-(√b²-4ac)÷2a

The AoS would be (-b÷2a) and yes, it's as simple as that. :)

For example:

Find the roots and the AoS of 4x²-6x-2=0 using the quadratic formula.

It's typically a good idea to write a=_, b=_, c=_, so you could have it for reference.

a=4, b=-6, and c=-2

1)x=(-b÷2a) +/- (√b²-4ac)÷2a --Starting Equation

2)x=(-(-6)÷2(4))+/-(√(-6)²-4(4)(-2))÷2(4) --Substitute in the numbers

3)x= (6÷8)+/-(√36+32)÷8 --Simplify

4)x=(6÷8)+/-(√68)÷8 --Combine like terms

5)x=(6÷8)+/-(2√17)÷8 --Simplify the term under the root sign, if possible.(68÷4=17, 8÷4=2)

AoS= x=6÷8=3÷4

Root#1:x=(6÷8)+(√17)÷4 Divide the denominator (8) by the coefficient (2)
and;
Root#2:x=(6÷8)-(√17)÷4

Feel free to ask any questions/comment if needed, and the next person to blog is Lisaaaa. :)

Quadratic Formula

Sorry I didn't get this done earlier, sorry for having a life now.

We learned how to change the formula ax² + bx + c to look like x = -b +/- √b² - 4ac ÷ 2a

Here's how it was done.

1. divide the equation by "a" -----> x² + (b÷a) x + c÷a

2. add (b÷2a)² to both sides and move c÷a (also divide (b÷a) x by 2) ------> x² + (b÷2a) x + (b÷2a)² = (b÷2a)² - c÷a

3. factor, move (b÷a) x to the other side of the equation, and multiply the numerator and denomenator of c÷a by 4a <- greatest common denominator ----> (x + b÷2a)² = -b²÷4a² - 4ac÷4a²

4. Simplify by combining -b²÷4a² - 4ac÷4a² so there is only one written denominator.

Now you have to square root the entire equation. Because you do not know if your result will be positive or negative, you put both options in when doing so. Doing that will result in this end equation: x = -b +/- √b² - 4ac ÷ 2a

Now you will use this formula to find the roots of the equation. This is an easier method to use. When you are given an equation (such as 3x² - 8x + 4) you substitute the numbers into the equation x = -b +/- √b² - 4ac ÷ 2a . Now, if you're wondering what numbers go where, remember the starting formula (ax² + bx + c) for finding the numbers. In this case, a = 3, b = -8, and c = 4. Put those into the new equation (x = -b +/- √b² - 4ac ÷ 2a) and solve.

Note: I know this might be hard to understand, just think of fractions when you have the ÷ sign, I couldn't find an easier way to make it look like its supposed to when you write it out.




Next blogger: CHRIS x)

Imaginary Numbers

Today in class we learn about imaginary numbers. Imaginary numbers are number that you can write down, but they're not possible to calculate.

Examples of imaginary numbers: √-2, √-4, 0.999999...., and etc...

Here are some things that you need to know:
Properties of " i ": i = √-1 and i² = -1 (note: i is not a variable like x or y, it never changes)

How to simplify:
ex. 1)
i² x i³ = (-1)i³ = (-1)(i²)i¹
^ here you change the i²s into (-1)
= (-1)(-1)i
^ and next you multiply the (-1)s
= i

ex.2)
5 + √-8 = 5 + √(-1)(8)
^ first you factor the (-1) out of the (-8)
= 5 + √-1 √8
^ next you square root the (-1), since the √-1 is i (reminder i = √-1)
= 5 + i√4 x 2
^ finally you square root the 8
= 5 + 2i√2

ex.3)
√25 + √-16 = 5 + √(-1)(26)
^ basically you do the same like ex.2
= 5 + √-1 √16
= 5 + 4i

ex.4)
(2i + 3)(i² - 5)
^ first you distribute the factors
= 2i³ - 10i + 3i² - 15
^ now you change the i²s into -1 (note: doing this first makes it easier)
= 2(i²)i - 10i = 3(-1) - 15
= 2(-1)i - 10i -3 -15
= -12i - 18

Trial questions:
1. 2i + 5 / i + 1 = (2i + 5) /(i + 1) x ( i - 1) /( i - 1)
^ in order to get rid of the i in the denominator you need to multiply it by the difference of squares
= 2(i²) - 2i + 5i - 5 /(i²) - i - 1 + 1
^ the (- i) and the (+ i) cancel each other
= 2(-1)- 2i + 5i - 5 / (-1) - 1
= -7 + 3i /-2

2. (3i² - 4)⁴ = (3(-1) - 4)⁴
^ the easiest thing to do is to change the i² into -1 and do the rest
= (-7)⁴
= 2401

That is all we did for half the class and the other half is our homework, well that is basically what we did all class.

Homework: Complex Numbers worksheets (all of the first page and the rest are even numbers questions)
Due: Thursday, March 15, 2007

Next blogger is Kirsty

Solving for x by completing the square

Sorry I didn't get a chance to blog last night so this is for yesterdays class, March 12, 2007.

For the first part of class, we looked over our previous test (Trigonometry) and checked for the correct answers amongst our peers. Then we went over how to solve for "x" by completing the square, rather than factoring.

Here are some examples:
1. x² - 4x +2 =0
Like in the Quad Functions unit, we follow the same concept when completing the square. You first focus on the "x² - 4x" and try to find the third term by taking the "4x" divide it by "2" and then square it. Then balance the equation by subtracting the "4" on the outside of the brackets.
( x² - 4x + 4 ) + 2 - 4 = 0
Now you can find the perfect square (x-2)² and put together the like terms ( 2 and -4)
( x - 2 )² -2 = 0
Basically all that's left to do, is to simply isolate "x"
( x - 2 )² = 2
x - 2 = ( + or - ) (the square root of) 2
x = 2 (+ or - ) (the square root of) 2
* sorry i don't know how to make the square root sign.

In this case, as in most, there are 2 roots:
x = 2 + ( the square root of ) 2 ( here, the square root of 2 is positive)
or
x = 2 - ( the square root of ) 2 ( here, the square root of 2 is negative)

2. This example follows the same concept as the first.
x² + 8x - 45 = -36
x² + 8x - 9 = 0
( x² + 8x + 16 ) - 9 - 16 = 0
( x + 4 )² - 25 = 0
(x + 4)² = 25
x + 4 = ( + or - ) (the square root of) 25
x = ( + or - ) (the square root of) 25 - 4

The 2 roots are:
x = 5 - 4 = 1 ( here the square root of 25 was positive, therefore it equals +5)
or
x = -5 -4 = 9 ( here the square root of 25 is negative, therefore it equals -5)

So this is basically what we did yesterday, and the blog for today is Shelly.

Untitled

First of all, I'm gonna blog about my Trig Reflection since I haven't had the chance to go on the blog for so long. So yeah, the Trig Unit was quite a review for me just because I just had Precal 10 last semester. Almost everything was the same except some areas like the Unit Circle which I understood better this semester than the last. So it was a good unit for me. I like it now better than the first time I learned it.

The class was pretty much a work period. But first, Mr. Wat (Sorry, not sure how to spell your name), wrote a couple of equations on the board for people to go up there and factor just to refresh our memory:

Ex: x² - 1 = 0

(x -1)(x +1)

Then find its roots.To find the root of a factored equation, simply make one of the terms equal to zero (0) so that it is "= 0":

(x-1) > What number will you use to replace "x" to make this term equal to zero? The answer, is one (1). The equation then becomes (1-1)(x+1). The multiplication rule states that anything that you multiply with zero (0) will equal to zero(0). So there you go, you found the roots.

After awhile, Mr. Wat assigned us Exercise #13, numbers 1 - 6. So we just worked on the exercise for the rest of the class.

Next person to post is hmmm, is KIMberly!

How to use your graphing calculator ..

Okaaaay. Getting back into the groove of things, let's get this blog going again.

Today, our new unit. ALGEBRA

What we basically did today was learn four different ways to use our calculators to find the roots (zeros or x-intercepts, whichever you prefer) of an equation.

The equation we are given is : 6 = x^2 + 6x + 11 (quadratic function)
0 = x^2 + 6x + 5 (standard form of the function)
*this is the equation we will be using for ALL the examples*

TRACE METHOD
step 1) Graph the related equation (standard form of the function)
step 2) Press the " Y= " button, and enter the quadratic function.
step 3) Have your " WINDOW " settings set to:
Xmin = -9.4
Xmax = 9.4
Ymin = -9.4
Ymax = 9.4
step 4) Press, " GRAPH ", " TRACE " and find X when y = 6. Use the left and right arrow keys to trace the graph.

ZERO FUNCTION METHOD
step 1) Graph the standard form of the function, by pressing the " Y= " button, and then entering the standard form.
step 2) Press " 2nd " then " CALC " and select " 2: Zero "
step 3) Guess values for the left and right boundaries. It has to be about 2 units less than the point on the left side of the x-axis, and about 2 units more than the point on the right side of the x-axis.
step 4) Select guess.

EQUATION SOLVER METHOD
step 1) Press " MATH " and pick " 0: Solver "
step 2) Press the up arrow key to clear the equation on the screen, a new screen should appear.
step 3) Press " CLEAR " to erase the equation beside the " 0 = "
step 4) Enter the standard form beside " 0 = " and press " ENTER "
step 5) Enter a number next to " x = " that is less than a possible x-intercept. So, in this case, you'd enter -2, due to the fact that -1 > -2.
step 6) Press " ALPHA " then " ENTER. " Round your answer.

COMPARING METHOD
step 1) Graph the quadratic function, and also graph " y = 6 "
step 2) Find the intersection (where they meet) point of the graphs.
step 3) Repeat for the standard form and " y = 0 "
step 7) Repeat on the opposite side.

Note: Your zeros should all end up to be x = -1 and x = 5

I know, it's a little tricky, but trust me, try the equations on the last page of the handout, and you'll get the hang of it.

That's it for me, later.

- Jho

Trig Reflection

It's Jackie again. Sorry that this blog is a bajillion years late, I wasn't able to log onto blogger because my computer's broken at home and because Ms. Ingram's computer was doing something funky with my account. Anyways, I found this unit actually pretty straight-forward, there are just some aspects which were a lot more complicated like the CAST system. I mean, it's a simple theory to propose, but it takes a while to actually catch on. Other than that, this unit almost felt like a review from what we did with trig last year in regards to SOHCAHTOA and SINE/COSINE law. The one thing that I have left to do with this unit is the group project, however. Not exactly sure when we're starting this but, anyways, until next time. Stay fit and have fun.

my reflection

This unit wasn't too hard, it was almost easy. There are some parts of it that I'm not too sure about but I'll wing it. I found ambiguous triangles the easiest part of the whole unit. the hardest for me was the trig equations. Anything to do with factoring is my weak point. I still am confused when it comes to unit circles... oh well. Good luck on the test, i guess.

Trigonometry

Well unit 2 is almost over after a few hours... When we first began I was a bit lost, but then I missed two days of school because I had the flu, which wasn't a great time to miss classes. So for a while I had know idea what to do. Until we have to make the unit dictionary that is when I finally got the hang of it... Well that is all.

Trigonometry Unit

Mkaay, so about, 13 hours till the test on the whole unit, and here I am blogging. Thanks to Marianne and Ivanna who reminded me. (: .. kk, so back on topic. This unit was, not super hard, but it wasn't a breeze either. I like the ambiguous triangle cases. It was tricky to understand at first, but when he explained the second time with actual examples, it became more clear. The whole " CAST " concept is somewhat fuzzy, but I basically get the idea. I remember using it from last semester, even though we weren't exactly taught how to use it. Standard angles and the reference angles aren't so much of an issue with me. The one thing I'm worried about is blanking out, cause I feel fine with all of it now, but once test time comes, oh gosh, I hope I'm ready. Alright. That's my blog for the second unit of grade 11 pre-cal. Good luck on the test guys! (:

- Jho

Trigonometry

hi everyone this is my second blog for the semester. im blogging on trigonometry. trigonometry was kind of confusing. i sort of get some of it but with other stuff its like huh? but yeah i guess its kind of easier because i remember some of it from last sem. so yeah thats my blog for tonight. its getting pretty late so i dont really have anything else to say. so goodluck to everyone on the test tomorrow!

TRIGONOMETRY

It's Ivanna again. I thought this unit was quite difficult. I was lost 80% of the time and I don't understand how to use the CAST thing. I also don't understand why we had to learn about the Sin and Cos graphs. I did like the exercises this time though, and we had lots of time to work on it. I'm straight owned for the test tomorrow. Good luck everybody.

blogging on blogging #2

hi there...this is john,,for me the second unit is quite hard at first..but as days past..i easily getting it more clearer because what makes me confused is the way our prof started his lesson though its good for the student,but as I remembered that topic from my country..they gave us a lot of tricks for solving that trigonometry..so,i think i will be good enough to answer the question to our test tom..i'm just having problem in labeling the triangles and finding the angle of it and also on the sine and cosine law because we didn't deal so much in that topic back in my former school..so,good luck to everyone for the test tom on trigonometry..