So we have this Inequality here; x² - 2x - 8 ≥ 0 So we're gonna go around about solving it like this;
STEP 1; Determine the Zeros of the function and place on a number line.
x² - 2x - 8 ≥ 0 → (x - 4)(x + 2) ≥ 0 → Zeros x = 4, -2 Now as you see this little bugger here, pay attention to the sign → ≥ Based on the sign, you'll know what you're looking for. So for this sign, we're looking for a positive (+) region. So let's put it on this little dandy number line. The positive areas represents that, those areas are positive, and the negative area represents the area between those two numbers is negative. I'll show you how we figured out which region is positive/negative lower down.
As you notice the circles on this number line, there is a rule ; > < → OPEN CIRCLE ≥ ≤ → CLOSED CIRCLE Therefore, depending on the sign, the cirlce will be opened or closed.
STEP 2;Test a value from each region.
So lets test values from each region; Test region (-∞, -2] So let's go ahead and try it on 4, -2 Lets just throw out a big number like -46545489454498457313216548. I think that's big enough. So we go along and → -46545489454498457313216548 - 4 = -46545489454498457313216552. Hey! A negative number.
Now try it on -2. -46545489454498457313216548 - -2 = -46545489454498457313216546.
Hm, still a negative number. Well with those two negatives, you multiply them to get... ( - x - = +) a positive value. Therefore, you've found the region you're looking for.
Now let's test another region Test region [-2, 4] Lets pick a number like -1. Sure.
Testing... → -1 -4 = -5 →-1 + 2 = 1 (Since it was -2, I didn't want to bother going - -2.) So now we have ( - x + = -) Nope. Not the region we're looking for.
Let's test the last region now Test region [4, ∞) So I don't know, lets pick a number like, 1454589658421259. No, no let's go with something like 5.
Testing... → 5 -4 = 1 → 5+2 = 7
Hey, so now we have (+ x + = +). Found it. ^_^
THEREFORE, the solutions are (-∞, -2], [4,∞). Or you could do it the LONG interval notation way which would be ; X ≤ -2. X ≥ 4. DONE! Now let's make it a little scary.
-x² +3x +10 > 0. Oh no it's a negative sign, rip up your papers now! Oh now wait, there's an easier way than burning up your paper. Just get rid of the negative so you can factor. To do so, just divide every term by - . Just to let you know, every sign will change. Yes including the greater than sign.
→x² -3x -10 < style="color: rgb(255, 0, 0);">NOTE; KNOW THAT THE SIGN HAS CHANGED. Now we can factor this little bugger. It boils down to (x-5)(x+2) < style="color: rgb(0, 0, 0);">So since the sign is <, we're looking for a negative region. I think she might have just done that to mess with us, but the truth is you don't have to do any calculations! Just use 0. So let's try it.
→0-5 = -5 →0+2 = 2
So the interval notation for this is (-2, 5) or you could do it the long way→ x > -2, x <>which could easily be done this way→ -2<x < style="" face="arial" size="3"> So you could'vedone it in one whole shot! Therefore it would be negative two is less than x which is less than 5.
Reminder that we have two dreadful Yellow sheets. Now who should be the next scribe... Roxanne? No now, lets go with Skyline! ^_^ No wait. I have a better idea. Let's give the honors to Elven! No hard feelings? Haha
Note to self; Do not blog using firefox anymore.
Well I know THERE MUST'VE been something wrong in this blog, so somebody please point out a mistake if I have one.. and so on and so forth.
And to end it on a funny note (I hope I don't get introuble for this.);
Okay, today we learned about Inequalities. But First!
Here are somethings that you should know in this concept.
Know those signs!
Okay, now that you know those you can learn the lesson that was taught today.
Steps to Solving Inequalities (Graphically).
1.Graph the line that will represent the boundry line.
If the sign is > or <>
If the sign is >= or <= then the boundry line is solid.
[Sorry i dont know how to make that character on the key board and i was not gonna make some images just for that. Sorry!]
2.Test one point on each side of boundry to find area that fits the inequalitie.
Sounds confusing eh? Well thats why we have...
~~~~~~~~~~Examples~~~~~~~~~~
1. y > -2x + 4
**Note about Graphing ===> Make sure the equation(s) are in y = mx + b Form.**
Once you have your graph you much choice a point on either side of the line of boundry. Any point will be fine, just choose one. For our lesson we have choosen (10, 10) and (0, 0)
The math:
(10, 10)
10> -2(10) + 4
10> -20 + 4
10> -16
Works!
(0, 0)
0> -2(0) + 4
0> 0 + 4
0 > 4
Doesnt Work!
Tip from Mrs. Ingram: Try to use the origin if possible. It will make the math much easier for yourslef.
You really only need to do one side when solving inequalites. If one doesnt work then the inequalite is on the other side!
~~~~~~~~~~Intersections~~~~~~~~~~
I'll explain as i go along.
x + y > 6and2x-2y<>
Make the equations into y = mx + b
You see that?! If you divide your multiply both sides by a negative value then the sign changes! Either from Greater than to Less than or Vice versa.
Once you do that you can graph it. And pick two points on each side of both lines to fine the inequalitie.
0 > 0 + 6
0 > 6
Wont Work
0 > 0 -5/2
0 > -5/2
Work
After you do that shade the side where the inequalitie is found. Do this for both Equations. After you will have something that looks like this.
Pretty crappy graph dont you think? Oh well thats not it. The red area is where the 2x - 2y <> 's inequalite area nd the blue is where the x + y > 6 's inequalites are. What you need to know is the purple area(both area's overlap). Thats where both inequalites of both equations are. Since the question says
2x - 2y and x + y > 6
The areas that overlap is the answer.
That was confusing. A little word like "And" can make such a difference. What about "or" There's a solution for that to!
I'll skip all the algebra and stuff. By now you should know how it works, if you dont...then i dont know. Just remember that when the you multiply or divide by a negative then the sign ( <> ) changes. :P
x - 2y <5 color="#000000">.This will look like that when you do the algebra.y > x/2 - 5/2
2x + y > 3 .This will look like that when you do the algebra.y > -2x + 3
0 > 0 - 5/2
0 > -5/2
Works.
0 > 0 + 3
0 > 3
Wont Work.
Your Graph will look like...
Again i know that sucks, alot. But just read this and find out about those little words. "Or" Refers to the sides of the equations with the inequalities, not just where they meet up. Everything in the blue Boundry is the inequalitie.
Arg! I made the lines solid. You should know that the boundry lines should be dashed. I'm just to lazy to do anymore work.
So in today's class we learned about non-linear equations and this question was given to solve for: 1) y=x-1 y=x2-6x-9
^ There are TWOvariables so that means you should have TWO equations. To solve it, substitution is the best method to use since y=x-1 already.
Now plug inx-1 as the value for the second equation: x-1=x2-6x-9
Put everything on one side of the equation: 0=x2-7x+10
Factor ! : 0=(x-2)(x-5)
There are TWO values so plug in the values to one of the original equations: x=2 or 5
ORIGINAL EQUATION: y=x-1
x=2 y=2-1 y=1 --> (2,1)
or x=5 y=5-1 y=4 --> (5,4)
Solution: (2,1) or (5,4) NOTE: You will have two points where it intersects. Like so..
The second example was: 2) x2-y2=5 xy-6=0
Again substitution is going to be needed in order to solve the equation. Rewrite the second equation so that it equals to y: xy-6=0 y=-6/x
Now substitute that into the first equation: x2-y2=5 x2-(-6/x)2=5
Then eliminate the denominator: x2(x2-36/x2=5) x4-36=5x2
Factor it: x4-5x2-36=0 (x2-9)(x2+4)=0 (x+3)(x-3)(x+4)=0 x=3 or -3 x+4 = imaginary root (4 will be a negative number)
Now plug in the values of x to find the points of y: y=-6/3 y=-2 --> (3,-2) y=-6/-3 y=2 --> (-3,2)
Solution: (3,-2) or (-3,2)
NOTE: Once again two points where it intersects:
YES ! I'm done. I was gonna pick Richard to get him back but then he chose ME again soo now I have to pick the LUCKY person that wins to be the next scribe.. and that person is PAULO ! Congrats and have fun. I heard and saw on the shoutbox that you haven't blogged sooo there you go! =D
OHHHH yeaaah.. Don't forget to do your homework: Exercise 26 questions 1-8 & 14
Step 2: Take two of the three equations and try to eliminate one of the three variables.
Example: 1) x+y-z=2 2) x-2y+z=-1 +_________ 4) 2x-y = 1 Once you've eliminated one of the three variables, the new equation will become equation 4
Step 3: Now take the equation that you did not use in Step 2, and take one equation that you have used in Step2, then eliminate the same variable as in Step 2.
Example: 1)( x+y-z=2 ) 2 ----- 3) 3x+y-2z= 4
2x+2y-2z = 4 <--- 3x+y-2z=4 -_________ 5) -x +y = 0 Once you've eliminated the same variables that you did in step 2, the new equation will become equation 5
Step 4: Now that you have equations four and five you can either subtsitute or eliminate one of the two remaning variables to sove for the other variable. Example: I will be using the subsitute method 4) 2x-y = 1 5) -x + y = 0 ----> y = x
2x-y = 1 2x- x = 1 x = 1 <--- solved variable
y = x y = 1 <--- solved variable
Step 5 : now that you have solved two of the three variables you can now substitute the variables into one of the original equation to solve for the last variable.
Example: x= 1 y = 1 1) x+y-z=2
1+1-z=2 2- z = 2 - z = 0 z = 0 <--- solved variable
(x , y , z) (1 , 1 , 0)
You now have solved a three variable equation you should give your self a pat on the back.
The next scribe will be Eric.... oh wait he is not in this class SO the next scribe will be Francis ohh know he cant be he already scribed this month. the next scribe will be Rojuane
Today in class we learned how to do a 2 variable equation question. Don't worry about 3 variables we never got to that point yet.
There are 3 ways to solve 2 variable equations: 1. Graph Basically you just have to graph the 2 equations, but to graph it in your graphing calculator it must be in a format such as this: y = [input equation here]. After both equations are graphed look for the intersect of the two, so look for where the lines meet. If they don't intersect at all, then it's not valid.
2. Elimination 5x + 4y = 6 -3y - 2x =-1
1. First you must make sure the x-values, y-valuesand numbers are in the same column. 5x + 4y = 6 2x - 3y = -1
2. Now you can choose a value to get rid of, x-value or y-value? I choose y. It's time to eliminate the y-value. You can do so by multiplying all the terms in each equation by the y-value of the opposing equation. 3(5x + 4y = 6) 4(2x - 3y = -1) ========== 15x + 12y = 18 -8x - 12y = -4
3. It's time to eliminate. Add or subtract to do so. 15x + 12y = 18 -8x - 12y = -4 =========== 7x = 14 x = 2
4. Substitute the value back into the original equation to solve for the second variable. 5(2) + 4y = 6 10+ 4y = 6 4y = -4 y = -1
Solution: (2, -1)
Time to see if the equation works: -3(-1) - 2(-2) = -1 3 - 4 = -1 -1 = -1 Correct.
If you end up getting an end result of 0 = 0 then it is always true. Any (x, y) point that solves the first line also makes the second line true.
3. Substitution 4x + y = 1 2x - 3y = 4
1. Solve the equation with 1 variable. 4x + y = 1 y = -4x + 1
2. Substitute that new equation into the second equation's y-value, and solve. 2x - 3y = 4 2x - 3(-4x + 1) = 4 2x + 12x - 3 = 4 14x = 7 x = 1/2
Check: y = -4(1/2) + 1 y = -2 + 1 y = 1 Solution: ((1/2), 1)
Oh yes, next scribe. Since I'm pretty sure everybody already went once, I'll cycle it through again. Which means Richard is the next scribe. Just nobody pick me halfway through the cycle. Since I went just now.
Hi, everyone_ This is first time blogging, so im really excited -_- Today's lesson was not really a new stuff_ And also it was very short_ made it very easy for me; Ok, so here's the question'
Given points A(-1,3) B(0,5) C(-2,6) , Vertify △ABC is a right triangle.
Well, if you have no idea how to do it, you might choose to just draw it, but it won't vertify anything_ so we have to consider these characteristics of right triangle_
1)
This is Pythagorean theorem that can be used for one of the ways to solve this problem =) You first calculate the distances of each side and see if this works or not_
2)We can calculate the slopes of sides and see if two pairs of adjacent sides are negative recicprocals of each other. If they are, they must be perpendicular which gives us 90˚.
3)Hardest way, we can consider angle measurements. One must be 90˚. I don't even know how to do this.
So, let's do the easiest way and get all three of the sides.
Today's lesson was just this_ I'm kind of feel something is missing.
So I will do another question for bonus that was on today's Exercise 23 #1-7.
I choose #4!
Line l₁contains the points (x,3) and (-2,1). Lince l₁is perpendicular to l₂which contains the points (5,-2) and (1,-4). Find the value of x.
Here's how you do it, or , i did it_
ok, i guess my blog is over_ almost_ lol So.. do i have to pick someone? o well, Francis?
gah sorry i couldnt do my scirbe.. hah scibe sounds like some new testament thing. anyways goodbye my marks. am i suppose to pick a new scribe? if soo then i guess i'll pick uh yuna or something (cool name) ookey cya
I forgot to blog yesterday so i am doing it today. haha. Well this unit i didn't find that hard. only remembering the equations was difficult for me soo i think that i will be doing pretty well on the test.
Anyways, This unit wasn't that bad, which is suprising considering I wasn't at school for most of it. Sometimes the blogs were kinda hard to follow, like there was too much alternate commentary along the way, and certain parts werent explained thouroghly but with a combination of Wikipedia, Google, and the Blog, I managed fine enough -_-;
Hmm, Hardest part would probably be everything I wasn't AT school for. So Imaginary numbers, and Quadratic Formula. Thankfully I started understanding those once I got back from my time off, so wasn't really all that hard.
Easiest part of the unit would probably be the radical equations (fewest steps to solve usually XD)
Overall the unit wasn't so terrible. I'm sure it would have been better if I was there for most of it, but eh, nothing much I can do about that. Well, that wraps that up this blog so good luck on the test, and I hope you all did your math dictionaries right! ;D
mhmm.. omg its so late ha kept on forgetting to do thiss.. but anywayss.. mhmm this unit was probably much easy than the other two. everything seemed easier to follow not a lot of new things to be learned, most of it i remembered from last semester, which is gooodd (: i guesss i disliked rational equations, just because they're fractionss and cuz sometimes i can't figure out the LCD.. grrr..! lol but anywayss.. i hopee i do well on tomorrows test and goodluck to everyone elseee.. (: yayyy
This unit is somewhat hard. Probably because of the radicals and fractions stuff we did. And I also sometimes get confuse in the Zeros of equation part. For some reason, i misunderstand this part of the unit, especially when i started doing my dictionary about that lesson. I kinda forget the difference between factoring and completing the squares. I also had a hard time in the radical equation part. It's just that sometimes it tooks me sooo long to figure out the answer fo certain question. The Quadratic Formula part is I think somewhat easiser than the other lessons. I also finf the roots of equation pretty much easier. Well I think that's all I could say. All I know so far is that this unit is somewhat is easy and somewhat hard for sooo many reasons.
Oh yeah, this be my pre blog for the test tomorrow. MAN I'M SO HAPPY I LOGGED IN =) ..
Yeah anyways, I would say this unit was very, very interesting. Besides that fact, it was pretty difficult! Knowing the unit dealt with tons and tons of square root signs... And other things. Other than that, I would say I enjoyed learning this unit and going through it. Woo thumbs up kinda.
Algebra... I thought that this unit was "okay". I don't want to say it was nothing and not do a very good job again. Most of it was pretty straight forward. Nothing really seemed that hard. When I went back through my notes to do my dictionary, I was lost at some points, but then it later came back to me, so it's all good! Anyways, I must head out and do... homework.
Hello. Well this unit felt like it just flew by. It seems like just a blur. Haha. It was really quick, with lots of lessons taught in it. But it was all good, let's see if it's all good after the test ! :O Haha. Okay, I'm off to study! Wish me luck tomorrow! Haha.
This unit was quite lengthy. The thing that aggravated me the most about it was that I was cut off by Spring Break. Nothing that really difficult just that some questions required a "different" method of solving. Overall, nothing was a surprise. It was pretty much branching off from Grade 10 Pre-Calc but there were some new stuff like Quadratic Formula. I still can't believe I have to do a project on it though...
Good luck to everyone on the test except for Paul Santos.
I thought that this unit was quite simple at first..and then the radical and rational equations were introduced. I hated those in grade 10 pre cal and I hate them here, I always screw up in whenever it comes to these types of equations. The pre test that happened today wasn't any better..I failed it. Harhar...then I re-did them at home and got the right answers and slapped myself across the face and back the other way. The only part about this unit that is troublesome is all the stuff you have to memorize, since this unit was really long, and I even forgot about that sum and product of the roots on the pre test also. Me and my short term memory...anyways.....I'm praying that I won't mess up and forget my formulas for the test..since I've been doing pretty bad since we came back from spring break...*sigh*..Now time to go study. 0_o
Lets see, there wasn't nothing too difficult, just my brain never had enough energy due to being deprived from sleep. >.< Tight schedule and all so never really got time to think clearly but anyways, it was alright. I'm in that in between point where I like and don't like my mark. I just hope I do well on the test tomorrow. X(
ookay i don't wanna say it was an easy unit cause last time i said that, i didn't do so well on the test the day after. so i'm going to say i liked this unit. nothing was truly complicated and i think i shall do okay on the test. i'm going to use this unit to my advantage and boost up my frightening-ly low average for math. oh man i hope there's a formula sheet. good luck to me!
Algebra....whats there to say? This unit seemed so long! Maybe it was becaues of spring break? Maybe it was just me? Anyways this really was a long unit. I mean look at the Dictionary! I needed to use four sheets of paper for it! Thats two more topics than usual! Meh. This unit wasn't really hard just seemed really straight forward. I dont think that there was anything that gave me to much trouble. I messed up the pre-test but I should be alright for the actual test.
I thought I was going back to school today, but I couldn't I had to go back the hospital.. but nothing serious, im okay now. ANYWAYS, I heard that the pre-test wasn't hard. This unit was okay, there's a couple things to remember like formulas and such but everything else is just making such your doing the steps to solving an equation right. I like finding the roots and I didn't like the radicals. So, good luck everyone on the test tomorrow annnd I'll be there... PROMISE. bye now =D
All in all this unit didn't seem bad at all. I didn't find anything hard at all. I forgot somethings like formulas and all these equations and terms but after you get to know it and remember everything it wasn't very hard. The break made this unit easier because it gave my brain time to relax and catch up, so I guess that's what made everything about this unit easy. I expect to get a good mark unless I forget some things. It should be all okay though. BYE!
Alright, it is finally my turn to create a scribe post. I will try to make this post the best ever and easy for all of you guys to understand. I will create a new standard for scribe post. Okay let's start.
Key Terms
Factor - One of two or more expressions that are multiplied to get a product. Example: 4 X 5 = 20 The 4 and 5 are the factors. 1, 2, 4, 5, 10, 20 are all factors of 20.
Expression - a combination of numbers, operators, and grouping symbols (such as brackets) arranged in a meaningful way which can be evaluated. Example: (X - 1) (X - 1) is the expression.
Free Variable - a variable which the value held in that variable is changeable. Example: X X can have a value of anything. The value of X is changeable at any point.
Bound Variable - a variable that stays constant at all times. Example: ∏ (Pi) The value of Pi never changes. It will always be 3.14...
LCD - this stands for the least common denominator. The least common denominator is the least common multiple of the denominators of a set of fractions. Example: (2 / 5) + (6 / 5) = (8/5) or {(X + 4) / (X - 7)} + (2/6) In the first equation, the LCD is 5. The second equation the LCD is 6(X - 7)
Restricted Values: values that the final answer cannot be.
Quadratic Formula: A represents the first term. B represents the middle term. C represents the last term.
The Lesson
I think the simplest way for someone to learn this material is by getting you a step-by-step guide of the process.
The equation will be {6x / (x - 5)} = 2 + {6 / (2x - 2)} Step 1: Factoring (if possible)
Factoring is taking each expression to it's simplest form.
This equation will be factored to be {6x / (x - 5)} = 2 + {3 / (x - 1)}
Step 2: Find the LCD
Now that the denominators are in their simplest form, we must multiple each denominator.
(X - 5)(X - 1) = X² - 6X + 5
Step 3: Determining the restricted values
To find the restricted values, we take the simplified expressions above and solve for X.
X - 5 = 0 | X - 1 = 0 x = 5 X = 1
Therefore, the restricted values for this equation are 1 and 5.
Step 4: Multiply each term by the LCD
We must now get rid of the denominators. Multiply each numerator by the LCD.
Now the equation becomes: 6x² - 6 = 2x² - 12x + 12 + 3x -15 Match like terms: 6x² - 2x² + 12x - 3x - 6 - 12 + 15 = 0 Combine like terms: 4x² + 9x - 3 = 0
Step 5: Solve Algebraically
The final step is just to solve for X. Normally we would factor the equation and find x. In this case, the equation isn't factor-able. This means we will have to solve by using the quadratic formula.
X = - (9) +/- √(9)² + 4 (4) (-3) 2 (-3)
X = - 9 +/- √33 -6
X = - 9 + √33 and X = - 9 - √33 -6 -6 Summary
In this post, I give you the definition of some terms. Then, I showed you how rational equations are solves. Basically, if you follow the steps I can guarantee 90% of the time you will not get the wrong answer.
This unit was suppose to be the longest unit of the year. Though, it didn't feel that long. The previous unit felt longer than this. I felt like I didn't have too much trouble on anything we did in class or on the homework. Most of it was an extended version of stuff we learned in the 10th grade pre-calculus class.
Hi you guys! This is Nelsa. Wow, I haven't posted on a blog (for this site) in awhile so I'm kind of relearning everything. =) This is so early. Well I have a spare and I'm on the computer anyway so I thought, "Why not just do it now before I forget and lose marks again?"
I can't believe our test is tomorrow, why did this unit go by so fast? Or.. was that just me? Well anyways, hmm, algebra. Weell, I don't know. I guess this unit was okay. It was neither easy or hard. [And that's exactly what I think of algebra too, soo, hah, it all fits!]
There wasn't anything in this unit that I really liked and there wasn't anything that I really disliked. Well.. except for the whole, working with fractions thing, but that happens in every unit so that doesn't count. Overall I thought this unit was.. well, okay. And.. yeah, that's it! So uhm, good luck on the test you guys. =)
Let’s start with a simple equationTo solve this kind of radical equation there are two steps you must do step 1) is to get rid of the square roots, to do that you must square both side of the equation
step 2) is to solve for “x”Example 1
heres how to check plug in the x value
Example 2
Try this one! Example 3
For this one move the 3 on the other side first to isolate radicals on one side. Hint* don’t over complicate the problem
example 4
Now try this one
follow the same procedure step 1) is to get rid of the square roots, to do that you must square both side of the equation to avoid confusion On a separate sheet of paper solve the first side of equation.
After doing this plug this in the equation
Now move everything on the other side to isolate the radicals on one side of the equation
Now square the equation again to get rid of the radicals Solve for “x”
YES! FINALLY DONE! woot woot!! sorry if i took me forever to do this i had work so yea anyways for the next scribber i pick hmmmm (look at the list) THANG N. since ur the first one on the list ur the next scribber =D if theres any mistakes please point them out thanks=D...haha well i better get some sleep hahah it sooo late, night everyone!
and btw; homework = excercise 18 #'s 1-9
oh n wait thanks remy n cell for showing me how to blog LOL
is a rule ;
ok, i guess my blog is over_ almost_ lol So.. do i have to pick someone?
step 2) 
Example 1 
For this one move the 3 on the other side first to isolate radicals on one side.
example 4
Now try this one
follow the same procedure
to avoid confusion On a separate sheet of paper solve the first side of equation.
Now move everything on the other side to isolate the radicals on one side of the equation
Solve for “x” 
