Category Archives: Math 11

Week 6 – Pre Calc 11

October 13, 2018Math 11shaylyng2016

This week in Pre Calc 11 we learnt three different methods on how to solve Quadratic Equations. The first method is Factoring, it is the fastest and the easiest method and there are two differnt ways that you can factor a Quatratic Equation. The first one is the Grouping Method which is the one that I prefer or the second one is the Box Method, which is a much more visual way to factor the equations. When factoring using the Grouping Method the first step is to determine which two numbers can multiply to the third term and add to the second term. For example, in the equation $x^2+8x+15=0$ you have to find two numbers that multiply to $+15$ and two numbers that add to $+8.$ Those two numbers would be $+5$ and $+3.$ From there, you can determine what the two factors of the first equation would be: $(x+3)(x+5)=0.$ Once you have found the factors you can then solve for $x.$ In order to solve for $x$ you must first isolate it. $x=0-3$ $x=0-5$ which mean that $x=-3$ and $x=-5.$

The second method that we learned is Completing the Square. This method can be used to solve any Quadratic Equation especially those that can not be solved by factoring. When Completing the Square of a Quadratic Equation you can first verify that the equation is not factorable, for example: $x^2+6x-13=0$ is not factorable because there are not two numbers that multiply to $-13$ and add to $+6$ which means that you can use the Completing the Square method. Once you have verified that the equation is not factorable, you can then rewrite the equation leaving space to add in your zero pairs: b = zero pairs. Ex: $x^2+6x+b-b-13=0.$ To find the zero pairs, you divide the middle number by two and then you square it. Ex: $\frac {6}{2}=3$ and $3^2=9.$ Once you have your completed equation, which includes the zero pairs, $x^2+6x+9-9-13=0$ you can factor the first three terms and since the last two terms are like terms you can combine and simplify them as well. Ex: $(x+3)^2 -22=0.$ You are now solving to isolate $x,$ which means you would add $+22$ to both sides of the equation. Ex: $(x+3)^2 =22.$ From there you square root both sides. Ex: $x+3=+/-\sqrt 22.$ Next you subtract $3$ from both sides to give you your final answer. Ex: $x=-3+/-\sqrt 22.$

The third method that we learned is the Quadratic Formula. This method can also be used to solve any Quadratic Equation but it is a little more complex and there are more steps, leaving you more opportunities to make a mistake. To start you have to determine which numbers are a, b and c and then from there you can put the numbers into to formula to easily solve the equation.

I have included an example below that shows the detailed steps I would take to solve a Quadratic Equation using the Quadratic Formula.

Week 5 – Pre Calc 11

October 9, 2018Math 11shaylyng2016

This week in Pre Calc 11 we focused a lot on reviewing how to factor polynomial expressions. Mrs. Burton gave us a fun acronym to help us remember the steps to follow when figuring out how to factor the epressions: CDPEU.

Common: $7x+14 = 7(x+2)$

Difference of squares (Binomial): $x^2-25 = (x-5)(x+5)$

Pattern: $x^2+x+4$

Easy: $1x^2$

Ugly: $4x^2$

One thing that I learned this week is how to factor Ugly equations. These equations are harder and more complex than the equations that we learned in Grade 10. For example: $x^2+1.5x+0.5$ may look like a very daunting and “ugly” expression because of the decimals, however if you were to change the expression so that the numbers were fractions instead of decimals than it would make it much easier to solve. Ex. $x^2+\frac{3}{2}x+\frac{1}{2}$ . Once you have changed the decimals into fractions, you must put each term under a common denominator. Ex. $\frac{2}{2}x^2+\frac{3}{2}x+\frac{1}{2}$ . From there you follow the acronym CDPEU and you find what they have in common and take it out of the equation. Ex. $\frac{1}{2}(2x^2+3x+1)$ . Once you have removed what they have in common you may than finish factoring the expression using one of the four methods that we learned in class. Ex. $\frac{1}{2}(2x+1)(x+1)$ . Onve you have finished factoring the equation, you can always check using the acronym to make sure that it can not be factored any further.

I have included an example below that shows the steps I would take to factor another challenging “ugly” expression that starts using fractions and not deciamls.

Week 4 – Pre Calc 11

September 30, 2018Math 11shaylyng2016

One thing that I learned this week in Pre Calc 11 is how to multiply radicals. Multiplying radicals is very similar to multiplying fractions, as Mrs. Burton told us you “just do it.” As long as the radicand is in the simplest form, you can multiply the like terms together. Meaning you multiply the coefficients together and the radicands together. Ex. $(2\sqrt3) (3\sqrt5) = (2)(3)\sqrt(3)(5) = 6\sqrt15$ . Depending on the question you can also use distributive property or FOIL. Ex. $2\sqrt3 (6\sqrt3+5\sqrt3 -7\sqrt3)$ . To FOIL an equation it is easiest to do so when all of the numbers are in their simplest form such as in the example. You can then add or subtract any like terms, meaning terms with the same radicand $(= 2\sqrt3(4\sqrt3))$ and then you can FOIL the equation. For this equation, you would distribute $2\sqrt3$ into the rest of the equation $(=(2)(4)\sqrt(3)(3)) = 8\sqrt9)$ . Once you have distributed, you can check to see if the equation can be simplified anymore or added or subtracted anymore $(=8\sqrt9 = (8)(3) = 24)$ .

I have included an example below that shows the steps I would take to multiply a more challenging equation using radicals and distributive property.

Week 3 – Pre Calc 11

September 22, 2018Math 11shaylyng2016

One thing that I learned this week in Pre Calc 11 is how to solve absolute values. The definition of an absolute value is the principal square root of the square of a number. This definition explains that the number to the power of two, square rooted is equal to the absolute value. Ex. $\sqrt8^2 = |8| = 8$ . The absolute value will always be a positive number also known as the “principal square root”.

You can tell a number is an absolute value when it has absolute value bars around it: $|3|$ . These bars do not mean the same thing as parentheses or brackets. However, if there are absolute value bars with an equation inside, you must solve whats in-between the bars before being able to find the absolute value.

I have included an example below that shows the steps I would take to find the absolute value of an equation, where you must solve what is in-between the bars before finding the final answer.

Week 2 – Pre Calc 11

September 14, 2018Math 11shaylyng2016

One thing that I learned this week in Pre Calc 11 is how to calculate the sum of the terms in a Geometric Series. A Geometric Series is when you find the sum of the terms in a Geometric Sequence. To find the sum, you must find the first term, the common ratio and the number of terms in the series. For instance, if this was my Geometric Series: $-2+8-32+128$ than I would use the equation $S_n=\frac{a(r^n-1)}{r-1}$ to figure out the sum. To input the numbers needed into the equation, I would replace $S_n$ with $S_4$ because there are 4 terms in the series. I would also replace $a$ with $-2$ because that is the first term in the series. Since $r$ represents the common ratio I would replace it with $-4$ and $n$ reprsents the value of the fourth term I would input it as $128$ . From there I will be able to determine the sum of all 4 terms in this Geometric Series.

This week I also learned that I can enter an entire equation into my calculator and it is able to solve it completely, which means that after I have inputed the numbers into the equation, the calculator is able to find the answer, which takes less steps and time to solve the equation.

I have included an example below that shows the steps I would take in further detail and the equations I would use to find the sum of the terms in a Geometric Series.

Week 1 – My Arithmetic Sequence

September 10, 2018Math 11sequencesshaylyng2016

My Arithmetic Sequence: -2, 5, 12, 19, 26…

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

$t_{50}=-2+(50-1)(7)$

$t_{50}=-2+(49)(7)$

$t_{50}=-2+343$

$t_{50}=341$

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

$t_n=-2+(n-1)(7)$

$t_n=-2+7_n-7$

$t_n=7_n-9$

$S_n=\frac{n}{2}(t_1+t_n)$

$S_{50}=\frac{50}{2}(-2+341)$

$S_{50}=25(339)$

$S_{50}=8475$

Week 1 – Pre Calc 11

September 8, 2018Math 11shaylyng2016

One thing that I learned this week in Pre Calc 11 is how to find the number of terms in an Arithmetic Series, when the number of terms is unknown. An Arithmetic Series is the sum of terms in an Arithmetic Sequence. To find the number of terms in an Arithmetic Series you must use the first and last number of the series and the common difference in the sequence. For instance if this was my Arithmetic Series, 3+7+11+15…+43, then I would use those numbers and the equation $t_n$ = $t_1 + (n-1)(d)$ to fiqure out how many terms are in the series. I would replace $t_1$ with the first number in the series 3, (d) with the number 4 as my common difference and $t_n$ with the final number of the series 43. From there I will be able to solve the equation and figure out what n=? (n= the number of terms). Once I have the number of terms I can then find out the sum of the numbers in the series.

Below is an example showing the steps I would take and the equations I would use to find the number of terms in an Arithmetic Series.

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