Want to PASS Algebra? You better know this formula
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- Опубликовано: 14 окт 2024
- TabletClass Math:
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Math help with solving a quadratic equations using the quadratic formula. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at tcmathacademy....
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In high school, I aced algebra. In college, I aced calculus. But it's been over 50 years since those days and I've forgotten a LOT! Thanks for the review.
Same here About 45 years for me A lol though is coming back to me I could Take College Algebra again And could definitely do well And Take College level Trig at the same time I am going to audit Both of the classes this fall Being a senior citizen does has it’s advantages
I was going to call you a show-off but you redeemed yourself with the end of your comment. 😊
Same here. I remembered that formula once I saw it. Good review. I have also been reviewing Trig and Calculus. I also want to study Geometry again. And Boolean Algebra. I have forgotten too much.
This is fun.
Me, too. I had a lot of college math, ending with Advanced Enginering Math (DE), but now I have also forgotten almost all of it. Ouch.
You guys are so lucky I went to high school back in the seventies algebra was not a mandatory class.. I never had it I could have taken it I wish I did now I'm learning it at 61 it's harder at this age but I'm getting the hang of it slowly but surely
I’m an 8th grader doing pre algebra and I decided I’m going to teach myself algebra and calculus since the school doesn’t want to
Quite right, go for it, then study classical mechanics!
7th Grader and still here
Get it son! Don’t stop after differential/integral calculus. Multivariable/vector calculus is the coolest
Nahh im a 6th grader 💀💀💀💀
I think you kids have already realized that the teaching of reading, writing, and mathematics have fallen by the wayside, in favor of a pro Marxist agenda, where you were not taught critical thinking, but instead you were taught to assimilate a doctrine that will destroy the nation in which you live.
I haven't seen this in 25 years and was quickly reminded of the formula. To this day, I have yet to find a use for this formula. Doing construction, many calculations are used involving algebra, geometry and some trigonometry, but nothing yet. Oh I remember the days of beating my head against the text book trying to understand.
It's great when a coworker asks how much is 200mm and after a slight mental pause, respond with the answer is roughly 8" because they have to allow that much clearance for an electric fireplace. Checking on a conversion calculator, the answer is 7 7/8". That math is important also.
It all depends on your field. I use it all the time. The transfer function of a second order low pass filter is of the form: G(s)= 1/(as^2 + bs + c) The zeros of the denominator give the poles of the filter, and you need these to figure out what the values of L and C you need to implement the filter.
Hey, Erin, I also have never seen a use for quadratic equations. My teacher in ninth grade showed us how to do them with the “magic formula”, but never explained why, or what it was for. Ugh. I went all the way through Calculus and never had to use it again. I also don’t see why it is named Quadratic.
1/2.54=X/20, 2.54X=20, X=7.874. Converting .874 to a fraction is where I slowed down. When in school 70+years ago, I had a bunch of fractional/decimal equivalents memorized, but I was just now surprised to learn that I had forgotten 7/8. Well, I'm not going to bother relearning those things. I remember my 12 times tables, and that's enough. I also know Pi = 3.14/159/26/535/8979 (that's how I memorized them). I knew more, but I've forgotten them. Why did I learn past 3.14? Well, I read about this guy in "Ripley's Believe it or Not," so I wanted to see how I compared [NOT VERY WELL!!]
I have absolutely no use for quadratic equations at my age and in my situation, but you got me interested enough to look up the history of them. I do remember an algebra teacher that took up (wasted) one entire class period by showing on the blackboard how the quadratic equation was derived. Oh, I should mention that I found your presentation very clear and very useful.
I loved the thorough, step-by-step breakdown. As a math tutor, I’m definitely going to need it!
Great experience, thank you, it can help with heaps of students who do not understand in equations like this! Thank you once again. God bless your online class,
Love ya...you definitely are a BIG HELP...Thank you
This is all about the so-called abc-formula, very easy when you cannot factor a quadratic equation. It also contains the so-called Discriminant (b^2 - 4ac), which informs you about the 'fractionability' of a quadratic equation (if the D = 0 you can always fraction the equation) and about the kind of the solution ( 2 real numbers, one 'double' solution or also imaginary numbers as solution(s) ).
Your explanation is excelent. Wish I had you as a teacher at school. ( I'm 72 years old) I knew the quadratic formula but Imade so manny mistakes that the result was scrambled eggs in a tumbledryer.
Maybe you just weren’t paying attention.
@@Willowdog08 😂😂😂🤦🏽
Same here, just didn't understand it in my schooling days. I have no use for this now but have the satisfaction of understanding it😂
I think it might be Bertrand Russell who said something similar to: "Education is the art of turning the obvious into something almost incomprehensible."
I was utterly lost 40 years ago when I was trying to learn this. Nothing has changed.
12:32 you should stress, that you are looking for a number that can be square rooted, therefore not 2x6. In an Example such like a square root of 64, (2^2)(4^2) both can be square rooted and be 8^2, but 2x32 can not be squared for a full number. You should emphasize you're looking for a number that can be squared.
In germany it’s taught in a slightly different notation x^2 + px + q =0 solves as x = -(p/2) +/- sqrt((p/2)^2 -q). Learned this in about eighth grade, but have admit since i didn’t use it in a few years i got rusty. Personally i think the notation is slightly easier to remember than your version, but naturally that is all a question of personal opinion.
How many of your students understand what this quadratic equation represents and the relevance of this "answer", or more accurately in this case, these two answers? Without understanding what it is they are trying to solve for, do they realize the possibility of having two, one or even no solutions and why those possibilities exist? I found showing them visually what they are trying to solve for and what the possible results mean made for better understanding of the solutuons they end up with. I had them do a rough graph of their results. It made for a better overall grasp of what it is they're working with.
The Quadratic Formula: To the tune of Row Row Row your Boat: x equals minus b, plus or minus the square root of b squared minus four a c, all over two a.
Excellent teacher/instructor. 😊
Thank you so much. I really appreciate you helping me understand this formula! :)
Thanks....it's fun to get back into algebra..... it's been 50 years for me.
Great video. I needed the refresher.
Hello. Isn’t it easier to first calculate Delta = square b-4ac over 2.a. If Delta is strictly positive there are two solutions then calculate x1= -b-Square root of Delta and x2= -b+square root of Delta. Your formula cumulates all and I think it can lead to errors of calculation. I am French, 76 and very poor in maths!
It is correct that the (3±sqrt(3))/3 can't be simplified to 1±sqrt(3), but it CAN be simplified to 1±(sqrt(3)/3) by first splitting it into (3/3)±(sqrt(3)/3) then simplifying the 3/3 to 1.
The method shown here using the Quadratic Equation Formula is best because the question is fill-in-the-blank. However, if it were a multiple choice question as on the SAT test, the fastest, easiest method is Sums and Products.
Step 1: Simplify the quadratic equation into x² - 2x + 2/3 = 0 by dividing both sides by 3.
The sum of the roots is always the coefficient for x with the sign reversed. Here this is 2. The product of the roots is always the constant which here is 2/3.
Step 2. Among the 5 choices, determine which choice has roots equal to 2/3 when multiplied or 2 when added. If there are no duplicates, that's the right choice.
Here multiplying the root choices that contain square roots might be easier using the axiom (a + b) x (a - b) = a² - b². The numerator is 6 and the denominator is 9 which is equal to 2/3.
The Quadratic Equation Formula should be a last resort used only when all easier methods like factoring, completing the square and Sums and Products fail or are too difficult.
Thank you. Agreed!
Confused as hell 😂🤣😂🤣. Definitely have to rewatch.
I took this in school, heard the other kids say why do we need this, the teacher said you will! I'm 57, a retired pilot, degree in Geological sciences, minor in math, never used this outside of school! Can't say that I ever saw this pop in day to day activity.
Just because you never used it doesn’t mean it’s not important. For anyone who wants to be an engineer or study physics or higher level math etc, understanding this formula is important for higher level mathematics.
@@JM-md4ri I get what you're saying, for higher level math, but is it necessary for the regular student that is not going to be an engineer, physicist, or whatever. I was just trying to point out that sometimes I think with all the other studies a student should learn, never do for subjects that may not be needed. I would venture that learning to wash and clean your clothes, is required more than these math concepts for 80% of the students. Not trying to offend, just a discussion.
y = a(x-h)² + k where h = 1/2(-b/a) and k = -a (h)² + c. a = 3, b = -6 and c = 2 so lets find h and k.
h = 1/2(6/3), h = 1
k + -3(1)² + 2 = - 1
y = 3(x-1)² - 1. (prove it: 3(x - 1)(x-1) + 2. 3(x² - 2x + 1) - 1. does in fact give us 3x² - 6x + 2
solve for x
x = h ± √(-k/a)
x = 1 ± √(1/3)
x = 1 ± √(3)/3 or as john shows it x = (3 ± √(3)/3). The 3 before the + or minus term is also divided by 3 so we can "pull it apart" so it is really 3/3 ± √(3)/3
bonus round
the y intercept would be 2
the vertex of this parabola is the the set [1, -1]
and the x intercepts are [1 + √(3)/3 and 1 - √(3)/3 ]
I never could remember the clunky quadratic formula, but the vertex formula I could remember. Of course I only use it if I am sure the quadratic equation can't be solved by factoring.
I have a masters in engineering and to be honest I have never bothered with memorizing the quadratic formula. I find completing the square to be more intuitive and just as efficient as plugging all the values into the formula then reducing. Plus I never have to second guess my answer and whether I remembered the equation correctly when I haven’t used it in a while.
Completing the square does not work well when the two roots include square roots, as it does here.
Impressive detailed explanation. Very very helpful. Thanks
I think creating (a +- b)^2 out of thin air where a^2 matches the quadratic part and +- 2ab matches the linear part with, most times, adding or subtracting a constant for correction on the right side is much quicker. Additionally you avoid multiplication of "big" numbers, rounding errors and end up with simple root-expression. Here (after /3) the expression is (x-1)^2 = 1/3
As a teacher you might be interested in a slightly different approach to solving ANY simple quadratic.
From the geometry of the quadratic we know that the curve is symmetrical about a vertical axis, let us say the line x = m. The roots R1 and R2 will be equidistant from this line. Without loss of generality we can assume R2>R1. The values R1 and R2 will be equidistant from the line x = m such that
R1 = m - d. and R2 = m + d.
By definition of what we mean by ' roots ' we can assert that
(x - R1)*(x - R2 ) = 0 --> (x - (m - d))*( x - (m+d)) = 0
x^2 - (R1+R2)*x + R1*R2= 0 -> x^2 - (2*m)*x + (m-d)*(m+d) -> x^2 - 2*m*x + (m^2-d^2)
referring back to the general equation a*x^2 + b*x + c = 0
m=-- 2*b/a and d = sqrt (m^2 -c/a )
R1 := -2*b/a - sqrt((4*b^2/a^2)-c/a) and R2 = -2*b/a + sqrt((4*b^2/a^2)-c/a)
The last line is complicated so in practice it's easier to remember the previous line expressing m and d.
Of course there will always be students who prefer to learn things by rote, and will not concern themselves with proofs. For the more serious student, I think this is a simple one to follow and my private opinion is that the expressions for m and d are simple enough to remember.
Excellent !
@@digitalkittycat4274 Glad you liked this approach. I take it you agree it's simpler to remember m and d ?
@@crustyoldfart There is an explanation video by Dr. Peyam, that fits I think.
ruclips.net/video/WRl0SRSF3BU/видео.html
@@digitalkittycat4274 He really gets enthusiastic. But I wonder why when explaining a basic proof in algebra, he does not use algebra, and insists on demonstrating specific cases. I'm wondering what audience he's addressing.
@@crustyoldfart I am not a judge, but ........
I guess he wants to prove. To prove you need "facts with evidence". I guess examples are the evidence. so he proves the fact (theory/algebra) with examples. :-)
You are correct. He should have used algebraic proof in addition !!
You said put that into your long term memory. I'm 68 and STILL know it despite never using it since school. Will that do?
Thank you for the thorough explanation; just too bad others didn't watch the entire video would have eliminated half of the unnecessary comments.
More than half a lifetime ago, I was taking an Electronics II exam and I got myself into a Calc mess & tried to back out using a Fourier transform. Got to the point where I could use the Quadratic to solve. Then the professor put it up because another students asked for it. “I don’t know why you’d need this.” I let out an audible f-word.
“Mr. Donovan, could you contain your Irish?”
Math is hard.
I prefer 1 +/- 1/sqrt(3). or 1 +/- sqrt(3)/3. And since everyone should know the decimal value of sort(3), 1.732... , you can get the decimal solution without a calculator in sight.
Yes I didn’t like how he didn’t factor out the 3
Right!? I'm goijg why, why is he claiming we canr factor this, it makes no sense we wouldn't leave it in lowest terms!
It should perhaps have been mentioned that if the part SQR(b^2-4ac) is negative. The answer will be a complex number. And in that case it OK to write no solution. At least in entry level algebra classes. Because if you have not learned about complex number, you can not deal with them
Why didn't you reduce your fraction? The answer I get is: 1 +/- 1/sqrt(3) My Alerbra teacher would've taken off a point or two for not reducing.
Then she would have been wrong too
... You also have the " COMPLETING THE SQUARE " method : Given 3X^2 - 6X + 2 = 0 ... X^2 - 2X + 2/3 = 0 ... (X - 1)^2 - 1 + 2/3 = 0 ... (X - 1)^2 = 1/3 ... (1): X - 1 = SQRT(1/3) = SQRT(3)/3 or (2): X - 1 = - SQRT(1/3) = - SQRT(3)/3 ... X1 = 1 + SQRT(3)/3 or X2 = 1 - SQRT(3)/3 ... S = { 1 + SQRT(3)/3, 1 - SQRT(3)/3 } ...
My math teacher tought me many years ago the next factoring:
a2-a2 = a2-a2 (No mistake!)
Left and right can be factored as follows
a(a-a) = (a-a)(a+a)
Then left and right can be devided by (a-a):
a = (a+a)
The brackets are useless now, so we can eleminate them:
a = a+a
Let’s simplificate that:
a = 2a
Deviding this by a gives:
1 = 2
With this riddle he wanted to illustrate a methematical taboo
Lol
That’s brilliant to revisit quadratic equations. Can it be solved in this way?
(3X-2)x(X-1) that way solution is x=1 or 2/3
It's no secret. It's the Quadratic Formula. I first learned about it when enrolled in high school geometry.
the quadratic formula is good to know but this example is easily solved using complete the square.
You're a very good teacher!. Keep it up
I do not agree. There are far better ones. He is rather confusing and uses a lot of blah bah in between and his voice is far too monotonous.
A good teacher would have explained why the formula is correct. Now it is nothing more than a trick. And math is not about tricks.
I will review in order to understand your method. Thank you
My first step would be to subtract 2 from both sides. After that, I had no clue. I have not worked with the quadratic equation since June 1971(Junior in h.s.)
I have to say that whatever program you are using looks similar to a win 95 MS paint program and I LOVE IT. Simple and effective. What program are you using?
The quadratic formula can be derived by using the completing the square strategy.
Surely you should be starting with BOMDAS / BODMAS ie: Brackets, Of the Power, Multiplication, Division, Add and Subtract. Applying a formula before doing the above is overly complicated when it need not be.
NO! BODMAS only applies to people doing "sums". Algebra is mathematics and relying on BODMAS produces ambiguities and thus confusion. No user of mathematics would ever rely on BODMAS when writing a formula.
Bro no way, I am thrilled to try this for first time in 24 years of my life.
Awesome video!
MR. TabletClass Math thank you for a fantastic video.
Why don’t any courses in U.K. or USA teach math about finances, savings, credit cards, stock market investments, APR, inflation and so on. These are life skills that I only realised in my 30s!
I live in Canada and it's taught in grade 12.
Because these are not Mathematics, they are Economics.
@@NacGT4 It doesn't matter what it's classfied as, it should be taught.
@@liamwelsh5565 My point is that they are taught! You just have to realise the applicable skills that you know!
@@NacGT4 When u have people in their 20’s asking their parents how to do it, it’s not taught. The problem is parents don’t always know how to do it either. Merry Christmas!
Actually the correct answer is x = 1 +/- [1/3(3)^1/2]. You need to simply your answer one step further.
Why not include the significance ie the two points where the graph of the parabola crosses the x axis?
hi profesor could you tell me what is the program you use for writting math?
Where did you get the ax squared plus bx plus c = 0 thing from? We were at the quadratic then you started using that one what’s that and when do I know to use it?
my high school math teacher would NOT have given full credit if the answer had not been further simplified, and properly presented as an equation solved for x...
x = 1 ± (1/√3)
🤓
.
Good review after 20-40 years, trying to help granddaughter
😊
Ok I remember how to do this Though I almost made mist mistake as My Algebra 1 Teacher would say “Watch your signs “ I got a little too over confident But I caught it ..Doing Math is a progression You build on what you have already learned As you progress you will have yo previously learn to do the next problems Simply stated you need the skills you learned in Algebra 1 to be able to do Algebra 2 You will need those skills to do Trig Once you have mastered those skills comes Pre - Calculus or Calculus 1 And you used all those Skills to do A lot More Calculus So you take all those above mention math Classes…You need them to do Calculus…With Calculus The door is open the sky is the limit
ok so I looked this equation up on a calculator site since I got a different answer, and it is basically the same, but the 3+-, is 1+ or minus. (The sqrroot3 over 3 is the same)
You could just derive the equation by completing the square.
...on the standard form of a quadratic equation.
This is the standard presentation of the abc-formula. I have a better formula based on this one with a few steps that make you understand parabolas.
My formula: x1,2 = xtop PLUS/MINUS delta
With xtop = -b / 2a and delta = SQR (D) / 2a and D = b² - 4ac
STEP 1 DISCRIMINANT
First you work out the discriminant D = b² - 4ac
When D > 0 then there are 2 points of the parabola crossing the x-axis
When D = 0 then 1 point touching the x-axis (of course this is the top of the parabola)
When D < 0 then there are no points crossing the x-axis, the parabola is fully above or under the x-axis
In this case: D = (-6)² - 4.3.2 = 36 - 24 = 12 so there are 2 solutions for y = f(x) = 0
STEP 2 XTOP
Now we know there are 2 solutions for y=0 we need to calculate the x-coordinate of the top of the parabola: xtop = -b / 2a
This formula is part of the original abc-formula.
The xtop is very important because the vertical line through xtop is the axis of symmetry of the parabola, so every point of the parabola has the same distance to this axis of symmetry.
In this case: xtop = -(-6) / (2.3) = 6 / 6 = 1
STEP 3 DELTA
This symmetry is also given for the crossings of the parabola with the x-axis, so the distance of x1 to xtop is the same as x2 to xtop
The formula of this similar distance: delta = SQR(D) / 2a
In this case: delta = SQR(12) / (2.3) = SQR(2.2.3) / 6 = SQR(3) / 3
STEP 4 SOLUTIONS FOR Y=0
The modified abc-formula: x1,2 = xtop PLUS/MINUS delta
So x1 = xtop MINUS delta and x2 = xtop PLUS delta
In this case: x1 = 1 MINUS SQR(3)/3 = 3/3 - SQR(3)/3 and x2 = 1 PLUS SQR(3)/3 = 3/3 + SQR(3)/3
Very clear
No need for any so-called secret formula:
3x^2 - 6x + 2 = 0
3x^2 - 6x + 3 - 1 = 0
3 * ( x^2 - 2x + 1 ) - 1 = 0
3 * ( x - 1 )^2 - 1 = 0
( x - 1 )^2 = 1 / 3
x - 1 = + or - sqrt ( 1 / 3 )
x = 1 + or - sqrt ( 1 / 3 )
Actually, you can go a step further by separating this "final answer" into (3/3)±(√3/3) = 1±(√3/3). This is your true final answer.
Yesterday I misplaced my damn cellphone (couch cushions swallowed it) and it took me forever to find it. Yet I remembered this quadratic formula I learned 40 years ago and came up with the right answer.
Is there an algebraic formula that explains what’s going on here? :-)
My son is struggling with the same question a lot of us struggled with when we were teens in the 9th grade - why do I need all of this meaningless stuff! I casually stopped off at work one day and told my son "let's tig weld that dirt bike muffler bracket back together." As he watched me program our 6 axis welding robot and proceed to lay a perfect circular weld, he asked me how I did that. I showed him the math behind the Robot Studio code and the light bulb in his head lit up. I could have welded that bracket at home in 5 minutes but the trip to work was a great learning experience. I think teachers need to spend time showing kids what all "this stuff" can lead to in life in terms they can relate to as a teen - video game programs, building dirt bikes and trucks, snowboarding, etc. It will all come together some day because everything evolves around math. Yes let's make it a challenge but fun at the same time. This is not a mind blowing 3X+1 situation.
This is quadràtic equation..x= plus/minus the square root of b squared minus 4ac all over 2a.
Me without dragging out my algebra book. How do you get from the first formula to the formula in the box? It's been about 30 years and I have never used it. But I work in electrical work and never used it.
ax^2 + bx + c = 0, where a does not equal 0
4a(ax^2 + bx + c) = 4a(0)
4a^2*x^2 + 4abx + 4ac = 0
(2ax)^2 + 2(2ax)(b) + 4ac = 0
(2ax)^2 + 2(2ax)(b) = -4ac
(2ax)^2 + 2(2ax)(b) + b^2 - b^2 = -4ac
(2ax)^2 + 2(2ax)(b) + b^2 = b^2 - 4ac
(2ax + b)^2 = b^2 - 4ac
2ax + b = +-sqrt(b^2 - 4ac)
2ax = -b +- sqrt(b^2 - 4ac)
x = ( -b +- sqrt(b^2 - 4ac) / (2a)
I'm going to be 70 years old on New Years day and remember this formula very well from my school days in the UK. I have just one question, I cannot remember why (-6) squared is = 36 and not (-36)
I have never heard a “logical “explanation, it was just a rule…. Plus x plus = plus; plus x minus = minus ; minus x minus = plus………sort of like a double negative in language……”I can’t not do it” means I can do it.
(-6)² couldn't be -36 because 6x(-6)=(-36) and 6x(-6) can't be the same as (-6)x(-6); but basically, yes, you just have to accept that negative x negative = positive. [And I beat you just: I'm already 70 LOL]
a negative number squared results in a positive result, hence -6^2= (6-) x (-6) = +36
Think of the (-) as meaning "reverse the sign." If you have a positive number 6 and reverse the sign you get (-6). This is (-1)*6. Reverse the sign again and you get 6. This is (-1)*(-1)*6 = 6. So if you reverse the sign an even number of times the sign is unchanged. If you reverse it an odd number of times the sign is changed. So (-6)*(-6) = 36 -- the sign has been reversed twice. (-6)^3 = -216 because the sign has been changed three times, from positive to negative to positive. And so on. It is entirely logical and you do not have to just accept it, but you can reason it out. A useful way to see this is by using a number line, and there will be lots of videos doing this.
Plse you have to remember that the polarities multiplied can change
Example: - × - = +
- × + = -
+ × - = -
+ × + = +
That is why the polarities are so important to remember
thanks
You’re the boss my 🎩 thank you 😊
I've always remembered that formula from 44 years since I had to use it. And although I've used many parts of maths again in my life, it's never once even remotely been useful. And to be honest, once I'd gotten square root of 12, I can't see any gain from reducing it further, since either way you'd be reaching for the calculator.
Mainly because a calculator cannot tell you the value of √12 [or √3] - it can only give an approximation. The aim is simplification; the same reason we reduce 4/6 to 2/3.
cross cancel?? My high school teacher would go crazy if he heard this. He preferred, "the 2's will Divide out" .
b^(2) - 4ac = 12
I don’t know why you call this formula secret, and you spent so much time to do it.
Why is a quadratic formula import? Tell me where I would use it? I left school 60 years ago, trained as an electrical engineer and have never seen an application for it. I have asked other engineers and get, yes I remember that but no idea where you would need it.
I forgot the last step. Your 3 + or - sqrt3 / 3 had me so confused.
Hello everyone,
for the mathematical solution of the quadratic equation 3x² - 6x + 2 = 0, I have calculated three possible solutions in various postings: the completing the square, the abc formula, which is presented in the video, and the pq formula. In addition to these computational options, I know of three ways to solve quadratic equations graphically.
The first is to turn the quadratic equation into the function y = 3x² - 6x + 2, plot this function using an appropriate program and have the zeros displayed with an accuracy of one or two decimal places. The program I used gives the zeros (0.42 | 0) and (1.58 | 0). It is better not to try to draw the corresponding parabola freehand with paper and pencil. It will never be accurate enough for the results to be acceptable.
The second option is to get a parabola stencil. I have one that is standardized for coordinate systems where one unit on both axes is exactly one centimeter. So you have to sharpen your pencil and draw very precisely to get the right result. Ultimately, you have to rely on a specific scaling of the coordinate system.
What you then do is convert the quadratic equation to x²:
3x² - 6x + 2 = 0 | ‧ ⅓
x² - 2x + ⅔ = 0 | -(-2x + ⅔)
x² = 2x - ⅔
Both sides are viewed as functions:
y = x²
y = 2x - ⅔
The normal parabola y = x² is drawn into the coordinate system using the parabola stencil. The straight line y = 2x - ⅔ is drawn into the coordinate system with a ruler and pencil, using the y-axis intercept (0 | -⅔) (approximately between -0.6 cm and -0.7 cm on the y-axis) and the slope 2. I got the values 0.4 and 1.6 as the x-values of the intersection points of the line with the parabola.
The third method is the Carlyle circle. In this case, you are free to choose the scaling of the coordinate system. I took three centimeters for a unit on both axes. For the Carlyle circle you need the parameters p and q, as they are also used for the pq formula. In our case p = -2 and q = ⅔. Now draw the two points A(0 | 1) and B(-p | q) in the coordinate system. In our case, B has the coordinates (2 | ⅔). The center of the line segment AB is the center of the Carlyle circle. You can either measure this center point or determine it with compass and straight-edge construction. How the latter works should be known. Now the Carlyle circle can be drawn into the coordinate system using the compass. The places where the circle intersects the x-axis are the solutions to the quadratic equation. I got the x values 0.4 and 1.6 for x₁ and x₂.
For comparison: the mathematical solutions are 1 - √(⅓) and 1 + √(⅓). The approximation, for example using a calculator, gives the values x₁ ≈ 0.42265 and x₂ ≈1.57735.
Best regards
Marcus 😎
I literally don't even know how I got an A in Algebra when this seems so foreign to me. I don't remember any of it. I would literally need to start in Pre-Algebra again and work my way back to here.
What is simpler, your solution or 1 +/- sq root of 3 over 3? Or am I totally wrong? It's been a long time since algebra 🤣
Good evening, genius sir, working out this sum
Ans,x=(6+✓60)/6,(6-✓60)/6
I am 74 and you just did what my teachers told me the b is -b when with out brackets it should be (b) you are implying that (-b)
I think you should have also mentioned 'Completing the square as a solution', particularly as this is how the quadratic formula was originally derived! In fact, do this to get the solution - very easy.
I would also have written the answer as: x = 1 ± 1/√3, which is a very tidy way of expressing it.
No one likes a “smart ass”!
@@markdavis9990 My suggestion was perfectly legitimate - any student learning about the quadratic formula should be made aware how it was derived. You're welcome to express your opinion, but my comment was made with the best of intentions. I would hope your comment was meant as a jocular response.
I got the same answer by completing the square: x = 1 ± 1/√3 which simplifies to 1 ± √3/3 (I think). Is this the same value as the above answer: 3 ± √3/3 produced from the quadratic formula? Can 3 ± √3/3 be simplified further like what one gets by completing the square?
Having an engineering background I would agree . However, 'proper' mathematicians don't like surds (√3) in the denominator (as I have learned teaching higher level mathematics for the past 30 years). I also agree with some of the comments that the completing the square method should be shown to be complete. Having said all that any help to increase student's confidence and competence in what they perceive as the difficult parts is to be congratulated. Well done sir.
@@dhulem ...and you're surprised that you got the same answer???
Binomial eqn. Pentagorias theorem .
there is a factor to take into account.
And I got the pleasure, Pleiades, The Tri Days, Steps, De Mond, we don't put an E
If it takes 17 minutes to solve one problem..... I flunked already.
Not exactly a secret, it was first published by محمد بن موسى الخوارزمي Muḥammad ibn Musā al-Khwārazmi in the 9th century! We also get the word algebra from his book الكتاب المختصر في حساب الجبر والمقابلة al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr w'al-muqābala. His Romanised name was Algorithmi from which we get the word algorithm.
How would that look on a graph? Giving 2 is the Y?
Hello everyone,
In addition to the completing the square that I used in my first posting and the abc formula from the video (which I used in my second posting), the pq formula can also be used to solve this quadratic equation. It would then go like this:
3x² - 6x + 2 = 0 | ‧ ⅓
x² - 2x + ⅔ = 0
------
x₁ / x₂ = -(p/2) ± √((p/2)² - q)
p = -2
q = ⅔
------
x₁ = -((-2)/2) + √(((-(2/2))² - ⅔)
x₁ = 1 + √((-1)² - ⅔)
x₁ = 1 + √(1 - ⅔)
x₁ = 1 + √(⅓)
x₂ = 1 - √(⅓)
Best regards
Marcus 😎
Bro that is the formula which is taught in class 10th to students and this is easy and almost every student uses it to solve the equations. But still that was nice explanation.
Thank you 😊💓
When I taught Alg 1 thirty some years ago, I required my students to be able to derive the quadratic formula as a formal proof before they could receive their final grade. The most awful crying and moaning and gnashing of teeth imaginable. The students I ran into later in their life thanked me because the confidence they gained allowed them to attempt higher level math courses in college.
You're welcome guys!
still trying to understand why the answer is not given as 1 plus-minus (sqrt3)/3 thinking it could be just a preference for how the answer is to be presented?
This may sound like a stupid question but exactly how do you know to plug in the a-b-c values for the numbers in the original equation? 5:22. Why A squared? why not B squared since B is squared in the Quadratic equation.
Do you mean by A squared is that ax² ?? So if that, ax²+bx+c = 0, where a, b and c only replace coefficients or constant. In Quadratic Equation, 'c' is the constant because it is the term that has no "x". The 'a' is simply the leading coefficient of the polynomial.
In Cubic Equation, we have ax³+bx²+cx+d = 0 where 'a' is still the leading coefficient while 'd' is the constant.
On the other hand, that b² inside the radical is not as same as a² or ax² that you may be talking about.
√b² - 4ac is called discriminant. It is just a part of the Quadratic Formula itself. Should be noted also that Cubic, Quartic, Quintic..... Equations do not have this.
X=(√3+3)/3,(√3-3)/3
Just didn't get the last part of the work
When are quadratic equations used in real world applications?
By completing the square , my answer is 1+_square root of 3 all over 3. Is this correct? Tnx
1.57 and .75 if the square root of 3 is 1.732.
Other than wanting to pass an algebra class, what is the purpose of learning this process? Thanks
I’m in Accounting. So I don’t know first-hand. I assume engineers need to know this in order to build things.
@@andrewm6424 Thanks for clarifying that for me.
By adding 2 and getting zero---the previous value must be negative 2.