Category: Math 11

Week 6 / Quadratic Formula

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The Quadratic Formula is a well known formuala used to solve quadratic equations; a quadratic equation is an equation (=) that has a squareroot symbole. To solve these you can use a number of things, but my favourite way is the Quadratic Formula.

To start you must know the Quadratic Formula which is x= -b +or-\sqrt{(b)^2 -4ac} then all of that divided by 2a. The A, B, and C, they all are pieces of the equation  x^2-6x+4=0  a=x^2 b=-6x c=4. The letters are just to show you where to place the numbers in the equation.

After you have plugged in the numbers the rest is pretty sympol, you just solve after that.

Week 5 / The Difference of Squares; Factoring

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Difference of Squares / Factoring

To understand how to factor Difference of Squares you must first know what Difference of Squares means. Difference means in math “subraction.” Squares means two numbers will square into another number. Example ({3}^2 {3 x 3} makes 9)

Forthgoing, figuring out if something is a Difference of squares or not, here two examples.

Easy equations

First when factoring you must find what is common, I found that {5x}^2 is something i can pull out of the equation and then your left with {5x}^2 (3x-1). To figure out if its aDifference of squares you check if it is squared and has a negitive sign; so yes it is a difference of squares.

The second equation is much more different than the first equation, its a trinomial, but that doesnt make a huge differenence when it comes to factoring the equation. Since both equations are easy equations, just checking what is common is all we need to do. In the equation 4 is the only number that is common, and your left with 4({2p}^3 – {1p}^2 -1). Since that is all you can do now you must find out if its a difference of squares or not. You atuomatically know becuase the 2p is cubed. So it is not a difference of squares.

Week four blog post

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Simple equations solving for X (with radicals)

1st step- Always the first step to solving with radicals is to find restrictions. Restrictions are a way to check that your answer is right by looking at the restrictions. When the equation that only has a radical and a coefficient infront, the same restriction applies. (x \geq 0)

2nd step- If there is a coefficient infront of the radical, do the opposite and divide the seven out from the radical. But watch out what you do to one side you must do it to the other; when you divided the seven, you have to divide the seven by the fourty two.

3rd step- Now that you are left with 6 = \sqrt{2x} to get rid of a square root symbole, your going to have to do the oppostite of square rooting, powering! You have to power the square root symbol, but what you have to do to one side, you have to do to the other. That means 6 turns into 36 becuase 6 times 6 is thrity six, well \sqrt{2x} becomes just 2x.

4rth step- Now that your just left with 36 = 2x, to seperate the 2 from the x you must to the oppostite and divide both the 2x and the 36 by the number attached to the x, in turn leaving the x by itself. So x = 18.

Finally- Your not done yet! Just becuase you have the asnwer doesnt mean your finished, you still have to check to see if you answer mathces your restrictions, if it doesnt its wrong. x \geq 0 does match x=18. That mean you got the right answer.

Week 3 Square roots

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Changing entire radicals to a mixed radical is fairly simple, but is a very important concept.

Your first step is to find out what the number in side the radical divides into. Fifty four can become  \sqrt{2} & \sqrt{27}. (Which was from \sqrt{54})

Since two can’t be made smaller it is left alone. Next is to make \sqrt{27} smaller, \sqrt{27} can be turned into square root 3 and 9. So now you have \sqrt{2}, \sqrt{3},  \sqrt{9}, nine can become smaller again and turn into 3. Nine is a special number called a prime factor which are numbers that can be multiplied by the same number evenly twice. If you have a prime factor like nine it just becomes 3 instead of square root three.

Now that your left with 3 \sqrt{2}, \sqrt{3} you have to multiple the square root 2 and the 3 to make it one number. After words your left with the answer 3\sqrt{6}

Write the entire as a mixed radical

\sqrt{54}

\sqrt{54}  = \sqrt{2} *  \sqrt{27} ——>\sqrt{3} *\sqrt{9} *\sqrt{2} ——-> \sqrt{3} * \sqrt{2} * 3 = 3\sqrt{6}

Write the mixed as an entire radical

-2 ^3\sqrt{5}

= –^3\sqrt{8} * ^3\sqrt{5}

=^3\sqrt{40}

 

Arithmetic Sequence

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Sequence: -13 -10 -7 – ….. +62      Determine the sum of the arithmetic sequence

62 = -13 + (n-1) (3)

+13    +13       —–> 75 = (n-1) (3) —–> \frac{75}{3} = (n-1) (\frac{3}{3}) ——-> 25=n-1  =    n = 26

/prt 2   S_{26} =  \frac{26}{2} (-13+62)   —–>   S_{26} = 13 (49)    —->    S_{26} = 637

 

Week 2 Convergent / Divergent

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Convergent – Where an infinite geometric series that’s sum gets so small it is almost hardly changing the sum. There are two types of infinite geometric convergent series; the first series gets mutiplied by a number less than one, and gradually getting smaller and smaller, but never reaching 0. The second series, R is less than 0 but greater than -1, and makes the series gradually smaller almost reaching 0 dipping up and below the positive and negitive line. Convergent is easy to calculate, and can be using S = \frac{a}{1 - r}.

Ex: 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}…

——>   S = \frac{2}{1-1/2}    =     S = \frac{2}{1/2} = 4 ——->  S = 4

Divergent – Is the opposite of Convergent, divergent is an infinte geometric series that increases and becomes larger, so large that it becomes almost impossible to find the sum of the infinite series. Just like Convergent, there are two types of Divergent series that can apear; first is a series that increases constanly upward positively being mutiplied by a number grater than 1 2 + 4 + 8 + 16 + 32 + 64….; the second is one that dips into the red when multiplied by a number less than -1. 2 – 8 + 32 – 128 + 512….. It is impossible to calculate a infinite divergent geometric series, and can’t be used by the equation S = \frac{a}{1 - r}.

My Arithmetic Sequence

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Sequence: 31 + 35 + 39 + 43…..       Find t_{50} & S_{50}

t_n= t_1+(n-1)(d)

Answer/ t_{50} = 31+ (50-1)(4)

=  t_{50} = 31+ (49)(4)

=  t_{50} = 31+ 196   =  t_{50} = 227

——-> S_{50} = \frac{50}{2} (31+227)

=   S_{50} = 25 (258)   —>   S_{50} = 6450