For us to find the sum, we will use the Quad Formula:
Now we'll add
-b+√b²-4ac ¯¯¯¯¯¯¯¯¯ 2a
to
-b-√b²-4ac
¯¯¯¯¯¯¯¯¯¯ 2a
to get the formula or equation for the sum of roots. It should look like this:
and we're suppose to do this,
-b+-b + √b²-4ac -√b²-4ac ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 2a
and the answer should be,
-2b
¯¯¯¯¯¯
2a
and because we could simplify this, it should be
-b ¯¯
a
This answer is our sum of roots. So if we have 2x²+5x-4=0 and your asked to find the sum of the roots you'll just have to use the equation above to find it.
eg. 2x²+5x-4=0 → a=2, b=5, c=-4
Ok, first part is done now we move to the second part,
Finding the Product of the Roots
Of course, for us to know what the product of the roots is, we will still need to use the Quad Formula:
And because we're looking for the product that means we'll need to multiply this:
-b+√b²-4ac ¯¯¯¯¯¯¯¯¯
2a
to this:
-b-√b²-4ac ¯¯¯¯¯¯¯¯¯
2a
It should look like this:
then you should have something like this:
The answer is our product of roots. And if we are to find the product of roots from the first example. we'll just have to do this:
Now the third and final part of what we learned this morning.
Determine the equation for given roots.
We'll just need to remember this equation or formula to find the answer or the equation:
x² - sumx + product = 0
Example. Given the roots of 3 , 9
-first find the sum → 3 + 9 = 12
-then find the product → 3 x 9 = 27
-now that we know the sum and the product well just have to replace it with in the equation:
x² - sumx + product = 0 → x² - 9x + 27 = 0
that was pretty easy eh? lol
ooh yeah..
*from the equation ax² - bx + c = 0. These the quad form we usually use.
a, b, and c should be integer coeffecient meaning it should not be in fraction or decimal form.
example. ( ½ , ¾ )
sum = ½ + ¾
= ⅝
product = ½ x ¾
= ⅜
x² - sumx + product = 0 → x² - ⅝x + ⅜ = 0
because there is a fraction and i just mentioned a while ago that we cant have fraction we'll have to do this:
8 (x² - ⅝x + ⅜ = 0)
multpily everthing by 8 (the denominator), and the answer should be:
8x² -5x + 3 = 0
Finally, I'm finish. I think I covered everything. Wow it took me forever to finish it sigh* im finally done!! yeeheyy!! lol.. and yeah for the next scribe or iunno what they call it.. It will be the pretty girl behind me.. sophia.. i mean Ms. Paulette ------- ...... lol.. yeah I know you already knew that your next but goodluck friendship... love you!!
Ok.. now i need to finish doing some stuff and man doing this blog is kind of hard and I probably made some mistakes (Im kinda not in me today..lol) .. so uhmm if i had some mistakes, anyone can tell me and i'll try to correct it.. thank you very much pipol!! =P
No fancy introductions, just the words: Let's get started. From the quadratic formula ---> x = -b±√b²-4ac/2a the discriminant would be b²-4ac A discriminant could be used to find out the characteristics or nature of the roots of a quadratic function. I will show three different examples showing how the discriminant is a clue to the nature of a quadratic function's roots.
EXAMPLE #1: y = 4x²-4x+6 In this equation a=4 b=-4 c=6 then you substitute these values into the discriminant and it would look like so ... b²-4ac = (-4²) - 4(5)(2) = 16 - 40 = -24 The value of the discriminant is a negative, therefore, we know there will be an imagincary number. The quadratic formula would look like: x = -b±√-24/2a < --- This would result to TWO imaginary roots In simpler terms : b²-4ac <> there are twoimaginary roots
EXAMPLE #2: y = 6x²+5x-2 So, we state the values of a, b, and c a=6 b=5 c =-2 then we sub these values b²-4ac = (5²) - 4(6)(-2) = 25 - 24(-2) = 25 + 48 = 73 Since the value of the discriminant is a positive, it tells us that its quadratic function will have two real roots. The roots would look like: x = -b+√73/2a , x = -b-√73/2a So, if b²-4ac > 0 there will be two real roots.
EXAMPLE #3: y = 5x²-10x+5 State the values of a, b, and c a=5 b=-10 c=5 then substitute these values b²-4ac = (-10²) - 4(5)(5) = 100 - 20(5) = 100 - 100 = 0 The value of this particular discriminant is zero. Now, when zero is subtracted from or added to a number, the answer is still the same. so, it tells us that it will have only one root. It would look like: x = -b±√0/2a In other words: b²-4ac = 0 there is only one root
So, always remember: b²-4ac <> b²-4ac > 0 ---- two real roots b²-4ac = 0 ---- only one root
*relieved sigh* ... well i know this was pretty late but... had work tonight and i was trying to get the hang of all those numbers. LOL well that's it for me... glad i'm done. hm... who shall i pick as a the next scriber?
I PICK .... eh... my bestestestest friend ever.... REMYSHIRE! haha sorry buddy, but there was no one else; you and i both know kamz isn't gonna do it =) bwahaha you know i love you!
well, i'm out... my dad's home and he's kicking me off the computer. oh yeah and if i made any mistakes (cuz i tend to do that a lot ... >_<) please point them out... nicely ? please and thank you.
Hello classmates, I will be scribing today's lesson (14/03/07) on solving equations using the quadratic formula. Let us begin this madness...
First of all, we have to find the the quadratic formula! Oh what fun... so let us get started.
To begin our quest to find the quadratic formula, we must first know what the standard form of a quadratic is, which we all should know already, but for those who have forgotten here is what it is:
>>> ax² + bx + c = 0
Alrighty, now that we know what that is, we may now begin working our way to find that darn quadratic formula! So first, we must start by completing the square of that abomination, and if you have forgotten how to complete the square, then you did not pay attention during the complete the square lesson at all! So here is what should happen to the equation...
>>> x² + b/a x + c/a = 0 If you don't understand what I just did, all I did was divide everything by a. >>> x² + b/a x = -c/a It was necessary to move the "c/a" to the other side so that I can create the quadratic formula, mkay? >>> x² + b/a x + b²/4a² = b²/4a² - c/a I got the b²/4a² by dividing b/a by 2 and squaring it, you know? And to make sure that the equation is balanced, I added the b²/4a² to the other side of the equation too =D! >>> (x + b/2a)² = b²/4a² - 4ac/4a² The reason why the other side is all over 4a² now is because I took the common denominator to make it..err..easier..to..work with. Yeah...
Now that we've got that down, we now have to take the square root of both sides, you know the drill. >>> x + b/2a = ± √ b² - 4ac I / 2a >>> x = -b/2a ± √ b² - 4ac I / 2a
*drum rolls* Aaaaaaand..here's our quadratic formula!! >>> x = -b ± √ b² - 4ac I / 2a
*Wipes sweat off face* Phew, now that we've got that out of the way, I will now show you something using our old friend Parabola!
Getting the zeroes using that formula is simple, as we all know. So..
x = -b + √ b² - 4ac I / 2a = -b/2a + √ b² - 4ac I / 2a
x = -b - √ b² - 4ac I / 2a = -b/2a - √ b² - 4ac I / 2a
That's all there is, and just one more interesting thing, the "-b/2a" is the value of the axis of symmetry and it is also the "x" value of the vertex. Cool eh?
Okay, I am not done yet! I'm almost..but not quite! I will now show you an example using the quadratic formula. There's going to be numbers this time, so no more of that a, b, c stuff..well..sorta.
Let's solve this equation: >>> x² + 3x - 9 = 0
Since we've just started using the quadratic formula, it is a good habit to begin a question by first finding out what the a, b, c values are. If you're not following me, then to make it more understandable, look back at the quadratic formula and you will notice that there are a's, b's, and c's. So we need to find those values in order to substitute the values in so we can solve the zeroes or something. If you're not sure which number goes with which letter, then look back on the standard form of a quadratic.
>>> ax² + bx + c = 0
Looks like that equation we're trying to solve right? Meaning that.. >>> a = 1 >>> b = 3 >>> c = -9
Wahoo!! We got our values, now we can substitue them into the quadratic formula! Oh, and again, since we've just started using the formula, it is a good idea to write it down first before you start substituting things in, so that you can memorize it.
>>> x = -b ± √ b² - 4ac I / 2a >>> x = -(3) ± √ (3)² - 4(1)(-9) I / 2(1) >>> x = -3 ± √ 9 + 36 I / 2
>>> x = -3 ± √ 45 I / 2 Uh oh, there's no square root of 45! Well, we know what to do! We should all know that 9 is the biggest square root that can go into 45 and 9 x 5 = 45 so yarrr!
>>> x = -3 ± (√ 9 I) (√ 5 I) / 2 >>> x = -3 ± 3 √ 5 I / 2 Oh kay! Now we can find the roots >__>..
>>> x = -3 + 3 √ 5 I / 2 >>> x = -3 - 3 √ 5 I / 2
Yeah..and that's it! Pretty easy..M I RITE? Sure I am! Well now that that's over, remember these key things when using the quadratic formula. 1) Make sure to find your a, b, c values first. Makes life way easier. 2) Always write down the formula first! If you don't...you might forget it! 3) I don't know. ..This key thing is for you to decide! XD
Anyways, I'm done. I hate scribing and this took me over an hour to finish..rawr. Now that I'm finally done, it is time to choose the next scribe. *Looks at list and sees that mostly everyone has scribed already* Argh, I guess I'll choose....chrycel. Yeah..mkay.
Oh and btw, for those of you who could not attend class today, today's homework was Exercise 14, questions 1-6.
Howdy Class! It's now my turn to do a Blog. Now I will attempt to explain today's lesson in a easy to understand way.
Remember Simplifying Radicals back in Grade 10?
Well here's a little refresher:
For Example, if you have √200:
√200 = √100√2 = 10√2
So, what do you do if the radical is negative?
Well here's an example If you have √-72 you just simplify it like normal except you take out the negative one.
Here's what I mean: √-72 = √36√2√-1 = 6√2√-1
HOW CAN THIS BE? THE SQUARE ROOT OF -1 IS NOT POSSIBLE! Well... We're gonna pretend it is... and call it an IMAGINARY NUMBER represented by the letter i.
So what I'm saying is that i= √-1. So if i is the square root of -1, then i² must be -1. Now how do you suppose to find i³? It goes something like this: i³ = i² × i = -1 × i = -i
Remember Kids, when multiplying powers with the same base you ADD the exponents.
Now on to the next part: EQUATIONS WITH i
Ok, what if one day we're in class and then Ms. Ingram all of a all sudden throws this equation at us and says to put it into the simplest form in terms of i:
-5i(7i - 9)
WHAT DO YOU DO!? WHAT DO YOU DO!?
Don't worry... I will show you how to solve it.
STEP 1: MULTIPLY it out using distributive law. So now your equation should look like this: -35i²-45i
STEP 2: Substitute for the Values of i -35(-1)+45i =35+45i
That wasn't so hard but TRY THIS: (3i² - 4)^4 I really hope you're not thinking about multiplying it out like this: (3i² - 4)(3i² - 4)(3i² - 4)(3i² - 4) That would just be simply WRONG!
This is how you really solve it: (3i² - 4)^4 = (3(-1) - 4)^4 = (-3-4)^4 = (-7)^4 = -2401
There's one last thing I must show you folks and it is: RATIONALIZING EQUATIONS WITH i
Alright, here we go!
Sample Equation: 3i + 7 i - 2
STEP 1: Multiply the original fraction by the denominator with the sign switched over. For Example, if you have √3 + 3, you multiply it by √3 - 3. In this case, we will multiply it by i + 2.
3i + 7 (i + 2) i - 2 (i + 2)
= 3i + 6i + 7i + 14 i² - 4
= 3i² + 13i + 14 i² - 4
= 3(-1) + 13i + 14 (-1) - 4
= -3 + 14 + 13i - 5
= 11 + 13i -5
That pretty much wraps up my post. Hope you guys understand it, especially you Justus because you couldn't be in class today. Well, hope your hand heals back.
Peace out
WAIT! I almost forgot. The next scribe shall be... Kristina.
Since someone decided to elect me as the scribe, I must now write a post on the topic covered today in Maths class (Today being March 12th, 2007). I hope you find this post informative and helpful :D
> But this is where things turn to THE TWILIGHT ZONE! > We will now boggle your mind!
>> (x+4)² = 9
> We have now moved the 9 over to the other side of the =. Now, we will solve for x! > But to do that, we must get rid of that pesky ². So we square root both sides!
>> x + 4 = +/-√9
> (Heres what it would look like were you writing it out on a pencil or paper) > Now we're almost done!
>> x = -4 +/-√9
> But here comes the tricky part! > Because the square root of 9 can be + OR -, we get two different outcomes (which means two different values of x). So we have to find the x value for both (which gives us the roots!)
>>> x = -4 + 3 └> x = -1
> OR
>>> x = -4 - 3 └> x = -7
> SO WE CAN DEDUCE....
>> x = -7, -1
> Dont forget to put them in coordinate form:
>> (-7, 0) (-1, 0)
> IF YOU'RE IN DOUBT, CHECK WITH YOUR GRAPHING CALCULATOR! > Like so...
The equation...
And the intercepts!
> And that my good fellows, is how you use Square roots to find the roots of a parabola. > Mind you, the number on the outside will not always be a number with a whole square root. Sometimes the darn thing will have to be left as a square root, in which case you'd just leave your answer as something like this:
>> x = -4 -√9, -4 + √9 >> x = (-4 - √9, 0) (-4 + √9, 0)
> This is also acceptable, but only when it cannot be further simplified.
> But now we must progress into another level of equation, and solve those!
>> 3x² + 12x + 21 = 0
> That evil fiend, coefficient has siezed the x²! Lets get em!
>> x² + 4x + 7 = 0
> Take that! This time, we wont simply put the 3 outside as a coefficient again, we have to eradicate him completely! > Now we can proceed in peace!
I would also like to recommend that if one wishes to show something from their graphing calculator, they might want to use the TI Connect ScreenCapture tool. Its much easier to read and faster than a webcam I think ;D
Hey guy just dropping you a line now that I've done my surgery. Everything went fine, although my arm is completely frozen. Like, I mean completely frozen, I cant feel anything at all along my whole right arm, and I cant move my arm or hands or finers. When I feel my right hand it feels really really weird, Like I'm shaking someone elses hand lol. In other words they drugged me up pretty well :]
Anyways what I really wanted to make this blog entry for was to ask you guys to makes sure that you include the exercise page numbers and question numbers with your blogs so that I can at least stay somewhat caught up with the class and not totally forget everything ^_^.
Anyways that pretty much that sumes everything up, I should be back on monday the 19th depending on how things go. See everyone later :), and remember to please add the exercise info to your blogs. Thanks!
Justus
ps. This took forever to type using only my left hand lol. In light of the fact I did it left handed please ignore spelling errors :p
trigonometry... love the whole thing. kinda surprised that i understood it after that issue from last semester where i was completely LOST. i really liked working with the ambiguous triangles now and it wasn't that hard but it was a bit time-consuming. the unit circle i liked and was very easy. and i finally remembered how to factor again. HAHA! hm. there wasn't really anything that was too hard in this unit not unless you didn't do any of the exercises or didn't go to class. well i'm off to school. =)
I Know its late, my sister hoged the computer all night >.<
So anyways, Trigonometry. Thank goodness we're done. Like, seriously Ambigouhumongousemessedupincomplete triangles? I had a little trouble with it but as usual, I got used to it. A lot of the work was very time consuming and that was probally the most aggrivating. But this unit went by pretty fast. Good luck all.
wow why was everyone having trouble with ambiguous triangles.. you're all pathetic. jus joking but i too have minor difficulties with ambiguous, and still don't think i have the hang of it. Reference angles and all were piece a cake, yknow finding like 8 different degrees, twas long process, and annoying but it was easy.i still dont understand what cos sin and tan actually are yknow like you press a button on your calculator and it does it for you. i dont even know what it's doing to the number i put in there. eh guess it doesnt matter.. for now. all in all twas a short lesson, aand good luck to me for the test.
I find this unit much easier than the last one. I enjoy doing the related angle stuff because its pretty easy and I understand it quickly the part that i kind of got the hard time is the Sine and Cosine curves. Its kind of hard in some way. I also did not like the exercises because some of it are like sooo hard to get. Over all this unit is I think shorter and easier than the last one. And i think that's all i can say about this unit.. I still need to finish my assignments sooo this is it. BYe.
Well I could probably sum up this unit with a few words. Good riddens. Of course, that wouldn't make for very good marks now would it :D Anyways this unit was fairly easy (and I mean it this time.) but like everyone else, ambiguous triangles took some getting used too. However, it all turned out to be another logic thing and in the end I found them to be do-able, and more along the lines of time consuming, rather then difficult. Overall the unit went by pretty quickly, and easily, with the only point of difficulty being the infamous ambiguous triangle.
wow this unit went by so fast.. i like trigonometry, i enjoy looking for the related angles i think that was pretty easy, and its good that we didnt do that much of graphing in this unit, its also good that there are not that much changes happening in the graph when we change the equation...hhmmm anyways... wut do i find difficult? we juss like everybody else... the AMBIGUOUS triangle **beep** that one was very confusing for me but the exercises kind of help me on understanding it.. i think? haha well thats all i can say about this unit... sorrie if dis doesnt make sense im very sleepy right now haha well good luck on the test everyone XD
Alrighty, blogging again… that was so fast… Well anyways, trigonometry, I didn’t think this unit was all that hard. Just like everybody else, the ambiguous triangles were confusing to me at first, but after doing some questions and looking at them, it didn’t seem as hard. One thing that really stumped me (and still does) are unit circles. I know all about the reference angle and stuff, but the thing that I blogged about before was all confusing. I don't know what the point of it is. What the *bleep* is it suppose to do?! ARG! Anyways, I must go enjoy the world of studying. Oh, and btw, anytime Xuan!
Hi guys.. sorry for the late blog. This unit was better than the last unit. Less things to memorize, no complicated forumlas. I would have to agree with some people about the AMBIGOUS TRIANGLES, it was kinda hard. But, otherwise the exercises took some time and the unit went by quick. hmmm, so that's it. see yah folks. -rojuane
I thought that this unit was way easier than the first unit. The ambiguous triangle gave me a hard time at the beginning but after a few tries I got it. My take on this unit is that it was hard at first then it became alright.
Okay... This unit was...meh pretty short and pretty easy. The only thing that killed me was....Ambiguous Triangles. Yeah we all had problems with it. Me...i dont nkow how to find if it has two different angles? Let's see what else i can say to make this longer... Yeah everything else was pretty easy. OH! I dont really understand that unit circle thingy...whats it's used for? Aside from those two things the unit was perfectly short and sweet.
Hey there everyone! :) .. hmm, this unit was pretty quick, but then again, the other one seemed quick to me also. Maybe they are a bit quick. Haha. To me, this unit was alright, most of it was good, but there was a thing or two that took me a while to understand. I think the easiest part was figuring out the ambiguous triangles, because it was all basic, and mostly solved using Sine Law. Well.. I think I'll end this here because I still need to finish a part in my dictionary. Hehe. Oh, and I need to study. Thanks to Jasmin, or I would've forgotten about blogging before the test :S .. Bye
hmm.. this unit was 'so, so'? lol well yeah pretty much i guess. it wasn't as hard as i thought it would be. everything was pretty much easy to understand but i don't like ambiguous triangles~ i can't get them right?! -_- booooo they make me sad. anyways, i think the last unit was easier though. well hopefully i'll do well for tomorrow's test! =) yaaayyy *crossing fingers* toodles
This unit was on Trigonometry. Slightly more tricky than the previous unit. I found that question on ambiguous triangles are quite lengthy and required too many steps to solve. I'd say that the only difficult part of this unit would be ambiguous triangles. Pretty much everything else was easy to do and understand. This concludes my 2nd Pre-Test Blog.
This unit wasn't that much harder than the first unit since most of it was stuff that I learned last semester, just with a few additions here and there. The only part that I found a tad difficult was the ambiguous triangles. I just can't seem to draw the diagrams properly, and that was where I messed up in the practice test.
All in all, I thought this unit wasn't so bad and my feelings towards the test tomorrow are just "meh". Mkay, I'm done now.
uh oh, another test another blog. For this unit on trigonometry it has been alright. I felt there was way to many formulas to memorize. I still don't remember how to solve those dang ambiguous triangles, we should of had more time to study and learn them thoroughly. But, that's just my opinion. Other than that the unit was actually quite easy. Homework was time consuming though, with all those steps. Overall the unit was okay, not good, not bad, but just okay.
The unit has ended. Our brains have enlarged from the new knowledge we have gained. At least our teacher and parents hope so. And yet again it is time to blog for our test.
Alright, what have I learned during this unit? I learned how to find if a triangle is an ambiguous one, find related/reference angles, and many other things. This unit was quite basic. There was nothing tricky or cruel. Just straight forward information. I thought this unit was fun. I'm pretty sure I did well during this unit being easy as it is.
