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This problem deceived me so hard, am I stupid? Factoring & quadratic formula, Reddit r/Homeworkhelp
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- Опубликовано: 14 авг 2024
- How do we solve 2x^2+x=6? Let’s learn how to solve this quadratic equation by factoring and by using the quadratic equation! Original post on Reddit r/homeworkhelp: / xztzsmg4hf
Where did the quadratic formula come from? algebra basics
• You use the quadratic ...
How to factor a trinomial (5 easy ways!) 👉 • 5 easy ways of factori...
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#math #algebra #mathbasics
![Gunna - today i did good [Official Video]](http://i.ytimg.com/vi/ulGLk8TaXcI/mqdefault.jpg)
Mathematically, which of these numbers is the largest?
(A) Number of hours in a year
(B) Number of seconds in a day
(C) Number of days in a decade
(D) Number of minutes in a week
Answer here: ruclips.net/video/06NqtlmkPK0/видео.html
Not understanding something doesn't make you stupid, you just didn't know it yet. Especially if it's math, it can be pretty difficult sometimes...
Right! I learned quadric equation at 9th grade
@@narayandaschowdhury4425 my uncle taught it to me in the 7th grade
@@flaregod9563 claps for u buddy
@@flaregod9563 Well, if you managed to remember or use it at all the next 2 years, cool
imo there should be 2 different words for being stupid: for not knowing something
and for needing more time (or being unable to independent of efforts) to understand something
in everyday speech both are called stupid
I know it looks harder but I find the formula way simpler to use. It's the same solution for every quadratic question and you just swap the numbers. Once I memorized it and got practice everything flowed out smoother and I never got a question wrong again
Yes, best part is that it involves no guesswork, the formula just works :), but sometimes teachers will want you to use the other methods and factoring can be useful in other ways as well.
I'm from Slovakia, and we didn't even be taught to use first method (by factoring), we directly were taught quadratic formula
same
The quadratic formula is very useful but it is also important to learn the completing the square method. This method will be required for putting quadratic equations from standard form into vertex form and for putting circle equations from general form into standard form.
@@MuhDog Once you know the roots, you can use them as a way of completing the square. If you want this process to be more mechanical, you need to normalize the equation (i.e., set a to 1 by dividing by a) first.
I'm an upper year canadian engineering major and I watched this whole video. no idea why but Im glad these resources exist for others :) mathematical literacy is really important!
😂
Even if I already know how to solve the problems, I'm learning a lot about _how to teach/explain_ it!
I did the same thing. I know how to solve this. I'm 31 and haven't been tested on this part of math in a long time, but I am thinking how this might help my younger sister who is learning this type of stuff currently.
lmfao i'm a second year biomedical engineer and i have a dif eq exam in abt 50 minutes... i also have no idea why i watched this
I have a masters in mathematics and I watched this whole video. The I looked in the comments and saw that I had already watched this video before and left a comment.
I almost always just use the quadratic formula to factor a quadratic
Same
I always calculate the discriminant b^2 - 4ac to determine whether it factorises or not. If it does then the square root of the discriminant is a hint to what the factors are.
That's good, you should do that if you need to do it quick. But being able to derive the quadratic formula gives a much deeper understanding, and gives you a more flexible approach that doesn't rely so much on rote memory
Do you know the middle term splitting method?
@@athrv._. that’s one way to call the factorisation method
Its always slightly impressive that he can switch between markers with one hand
That was my big takeaway. I mean, I already knew the math, so when I noticed that he was writing in two different colors with the same hand, that became my focus for the rest of the video. He's slick, palming a second marker so that he can quickly switch like that.
...while explaining the math.
bprp literally stands for "black pen, red pen." Those are the colors he most often has in one hand.
oh my god hahaha i was just thinking the same thing... ive never seen a teacher that does that so seamlessly
Hey, I completely forgot about this! I just looked at it and thought it was -2. I'm glad I clicked on the video to check. It's been 16 years since I've thought about the quadratic formula. Still had the equation drilled into my brain lol. Great refresher!
I’m not trying to shame you or anything, but how do you get that? Bcz I just got out of high school so these formulas are still really fresh for me. Again, not trying to shame or disrespect you
@fabo-desu A lot happens in life after school and if you aren't using the knowledge and keeping those neurons freshly connected it can fade. I got my answer by solving it like a quick puzzle in my head. However, I always check to make sure I'm right because I know some things blur together. A good example is Pokémon Red Version. I can still recall where every Pokémon and invisible item is (even the Pokémon levels you can find in certain areas), but I recently got stuck for a couple of minutes in the Dark Cave without flash and ran into a couple unnecessary fights, which never happened a kid. Last time I played the Red version had been 20 years ago. Even though I remembered almost everything, there are pieces of memory that stay dormant until you have to re-access it.
Once he mentioned setting it up as a quadratic equation, it clicked back for me. I just haven't seen a quadratic equation since 10th grade for pure math problems. A bit of physics in college as well, but I only took a semester before going into the Navy for Electronics Technician. You don't really need to know quadratic equations or Calculus to take electrical measurements, replace resistors and capacitors, or work on computers. Hell, these days I do IT work. That's why I try to keep myself sharp when I see things like this. If I didn't, I would never think about it. Hope that makes sense! I was pretty on point with stuff from highschool until about 27ish. I'm 33 now.
That's pretty intuitive, you try to gor for integers, 1 is too smal, 2 gives you 8+2=6 so it's too big but then you see that just changing +2 to +(-2) makes it's right and it doesn't matter if the number is negative if it's in even power since - x - is + as well as + x + is a + too.
@@DarkDragonRusThat's exactly my thought process after seeing the preview-picture, but it only really works with relatively easy numbers, so I'm still happy to have seen the video to refresh a little bit of math-knowledge.
Same, same, 15 years since I've had to do a quadratic equation but I love maths and use probabilities and factors every single day of my life but going back to the hard stuff, really wrinkled my brain for a minute, feels good.
Original Question:
2x^2 + x = 6
You can either factor (if possible), or use the quadratic formula. Before we try the quadratic formula, let’s see if the equation factors.
Factor:
2x^2 + x = 6
2x^2 + x - 6 = 0
Because the leading coefficient is ≠ 1, we have to find a number that multiplies to be a * c:
a = 2, c = -6
2 * -6 = -12
And that adds to be = b:
b = 1
Those numbers give us 4 and -3 (4*-3=-12, 4-3=1)
(2x^2 + 4x) - 1(3x - 6) = 0
2x(x + 2) - 3(x + 2) = 0
(2x - 3) (x + 2) = 0
Solve for X:
2x - 3 = 0
2x = 3
x = (3/2)
Check:
2(3/2)^2 + (3/2) = 6
2(9/4) + (6/4) = 6
(18/4) + (6/4) = 6
(24/4) = 6
6 = 6 ✔️
(3/2) is a valid answer
Solve for X:
x + 2 = 0
x = -2
Check:
2(-2)^2 + (-2) = 6
2(4) - 2 = 6
8 - 2 = 6
6 = 6 ✔️
-2 is a valid answer
Final Answer:
x = (3/2), (-2)
This is easier than tik tok method because you do not have to guess. Only think of two numbers that thier product is ac ( here -12) and thier sum is b ( here is 1) . The answer in this case -3 & 4 and use this to split the factor of x . The good thing is that the order does not matter and it will work. I will use diffrent order
2x^2-3x + 4x-6=0
x(2x-3) + 2( 2x-3)=0
(2x-3)(x+2)=0
x =3/2 , x =-2
Still don't know why it's kadebthis difficult. Just say itvis equal to x^2 +0.5x = 3
= (x + 0.25)^2 - 0.25^2 = 3
= (x + 0.25)^2 = 3.0625
= x + 0.25 = sqrt( 3.0625) or - sqrt( 3.0625)
So x = 1.5 or x= - 2
Nice use of Chat GPT 😅
I just realized that having a 4 year old means i have roughly 10 years to relearn this stuff to help her with her math classes
I click these because I'm interested in the maths, then I get completely mesmerised by the effortless pen juggling. Every, single, time.
This channel is absolutely brilliant! I excelled at math throughout high school, did it at a University level to some degree and forgot how to solve something so basic. My mind was just substituting numbers but completely forgot I could go negative or fractions. Love your work!
I wish this channel had been around when I was taking these classes.
I am genuinely grateful there are people out there who love math and can do math. I'm not one of them. Godspeed, everyone else!
The way I used factor method is to calculate ac and guess its factors which sum to b.
In this example , I found ac = -12. I then tried to think of factors of -12 which adds up to b=1 . The factors were found to be 4 and -3.
2x^2 +x -6 =0
2x^2 +4x -3x -6 =0
2x (x +2) -3(x +2) =0
(2x -3)(x +2) =0
This is exactly how I solve it as well. Seems much simpler!
After using the "guess and check" method for years, I learned the method @Dumpy332 outlined here. I HATE guess and check for ANY math. An algorithm that always works is MUCH better. Having said that, I had a student that always used the quadratic formula no matter what, and I couldn't talk her out of it. She's a lawyer now so I guess it worked for her :) As an aside, I always tell my students that the difference between "hard" and "easy" is that you know how to do the easy stuff but don't know how to do the hard stuff. At least I've found that true for many, many things, even beyond math.
Other way to solve is by completing the square which has no guesswork and is basis of quadratic formula:
2x^2 +x -6 =0
x^2 + x/2 -3 =0
x^2 + 2(1/4)x + 1/16 - 1/16 - 3 =0
(x + 1/4)^2 - 49/16 =0
(x + 1/4)^2 - (7/4)^2 =0
(x + 1/4 + 7/4) (x + 1/4 - 7/4) = 0
(x + 2)( x - 3/2) =0
@@Dumpy332u divide 2 side to 2 xD. In my country we have one person trying to solve 1 math ex with more than 20 different way lmao. Math is crazy fr
Shortcut: the even term can't be the lower one. If it was, the middle term would be even. If the product is irreducible, so are the terms. Whenever a and c have a common factor, you should put it in the x-term of one and the 1-term of the other. (If it's in a power, you shouldn't split it. This only applies to common factors. If there are multiple, you can put one in a fixed place and consider cases for the others. Or you can use the formula)
I love seeing the fundamentals as well as the hard stuff that's difficult to explain. If for no other reason so I won't forget how the basics are done.
Keep up the great work BPRP.
I always found the cross method to be my go to for solving quadratics, wherein you draw a little X and on the top you multiply the constants a by c, and on the bottom you put b. Then you ask yourself what two numbers multiply to give you the top, and at the same time what two numbers add up to give you the bottom? in this case, we would have 2x^2+x-6 = 0, so the top would be -12 and the bottom would be 1. You can just go through all the various factors of 12 in your head (sometimes if the number is large it can take a bit), but the only combination that works would be -3 and 4 (-3*4=-12, -3+4=1), and that would be your -3x+ 4x terms, which you would then group and factor to give you the two factors.
In a similar vein, I did a similar process for FOILing (which I realize years later is actually a geometric method for showing it lol), where you draw yourself a "railroad" or "box" of the two factors and you multiply the terms in the little box by column and row, and add them up
unfortunately the cross method doesn't work with all quadratics, which is why i find completing the square so much easier - it not only tells you the turning point of a quadratic, but also allows you to solve it and find both real and imaginary solutions - much like the quadratic formula.
@@hardstuck6200The completing square method works on everything cause they aren't dependent on the discriminent while for factoring with the cross method requires the discriminent to be less than 0
@@anujaamarasekara9513 i know thats what i said
@@hardstuck6200 fair enough, my mistake
When I learned the quadratic formula, my teacher had us sing it to the tune of "Pop Goes the Weasel." As silly as it may have felt at the time, I guarantee none of us will ever forget "x = -b plus or minus the square root of b² - 4ac, all over 2a"!
I'm a 49 year old engineer and I watched the whole video. I automatically do the quadratic formula since it's just easier for me.
I always have to solve for the quadratic formula because I can never remember it, so I default to factoring. I'm a 32-year-old engineer 😅
√49😂
whenever I see videos where teachers try and explain fundamental concepts I always notice things that they could improve on so their explanation is clearer and easier to comprehend, however here I think everything you said and more importantly HOW you said it was perfection. Something I hope I can also do in the near future
Thank you!
@@bprpmathbasicsI agree this was a great video, but one small thing I think could've been helpful is a short explanation of why you set to 0 in factoring to get answers.
@@benrosenthal4273For me that’s where the graphical method comes in: I remembered all of a sudden that this is a parabola and what’s “0” is the value of y. The two solutions for x are where the parabola hits the x axis on each side on the way down. Also, other properties of the curve like how many inflection points or its shape drive the equation. I am very good at manipulating images and visually represented math/geometric concepts in my head…not as great with handling the nitty gritty number crunching, or dealing with garbage teachers throw out without them plotting out what’s going on visually first.
What was the deceiving thing about the formula? Is there something to trip over?
I thought the answer is 1.5 and other people seem to think the answer is -2, so I probably hit the trap, whatever it is. But I don't see how 1.5 could be wrong.
@@godlyvex5543 those are the two answers, they are both correct
2x²+x-6=0
We know that parabola are symmetric.
So, for a well chosen "u", we have: f(u-x)=f(u+x)
a(u-x)²+b(u-x)+c=a(u+x)²+b(u+x)+c
-2aux-bx=2aux+bx (skipped same terms)
-4aux-2bx=0
Let set x=1 to solve for u and simplify.
2au+b=0
u=-b/2a
So, -b/2a is where our vertical axis of symmetry pass.
Now, our solutions have to be on either side of -b/2a or -1/4
2x²+x-6=0 => x²+x/2-3=0
(x-a)(x-b)=x²-(a+b)x+ab
ab=-3
a+b=-1/2
(-1/4+t)(-1/4-t)=-3
1/16-t²=-3
t²-1/16=3
t²=49/16
t=7/4
-1/4+7/4=6/4=1/2
-1/4-7/4=-8/4=-2
I am in 8th grade and I did it exactly how you did, but just a correction.
7/4-1/4=6/4=3/2 not 1/2.
@@TheAwesomeAnan That was the most complex part ^^
@@programaths lol
@@TheAwesomeAnandamn in my country 8th grade has pythogoras theorem and comparing numbers...
Like ratio and %
We learnt parabola and symmetric functions in 11th
@@epikherolol8189Same in Belgium and we are the 13th in PISA ranking.
We can solve problems like "By operating only on denominator, augment 5/7 by its 2/3".
Ratio are much more complicated than quadratics.
That's because you can get away with one formula for quadratics, or using the above or completing the square.
On the other side, proportions requires a variety of concepts and more adaptability.
5 machines do 5 widgets in 5 minutes, how many times do 10 machines to do 10 widgets?
To make a glass of grenadine, you need 1 volume of sirup and 7 volume of water. How much sirup do you need for 1L?
One person takea 30 minutes to do a task. Another person takes 40 minutes to do a task. If they work toghether, how long would it take for them to do the same task?
Convert 100kph in mps.
A person walking at 5kph on a 100m long conveyor belt. He takes 5 minutss to go back an forth. How fast is the conveyor belt?
And also, look for a video about how to calculate base tax in Belgium that I posted on my channel.
You can see that each problem need a different way to approach it and I didn't even touch the geometric aspect. (Thales and similar triangles)
We push quadratics down the line to do them at the same time as more advanced algebra when we study function (limits, assymptotes, derivatives...)
The quadratic formula is very interesting, you can analyse what happens in the square root with different values and it will reveal some amazing things :)
What "amazing things" do you mean?
@@bjornfeuerbacher5514
Positive: Two Reel Solution
Zero: One Reel Solution
Negative: Two Complex Solutions.
@@Ribulose15diphosphat And what's amazing about that? That's standard, quite basic knowledge.
@@bjornfeuerbacher5514might be amazing to someone who didn’t know it before. There was a time where you didn’t know it, and I bet you would’ve found it cool when you learned it. So unnecessarily condescending!
@@dimwitted-fool Probably I misinterpreted him, I thought he meant that this would be amazing for BlackpenRedpen - and I think we agree that BlackpenRedpen has already known that for a long time. ;)
I only ever knew to use the quadratic formula. It never occurred to me that you could also solve by factoring.
Tbh looking at this here I'd say the quadratic formula would be easier and less likely to lead to mistakes, especially in real world scenarios.
I would have loved to have had you as my math teacher in college. I'm retired and don't need to do any math, but I enjoy your lessons.
We only learned the Quadratic formula in school, but I had a friend in that class that did Kumon and there they're also taught quadratic factoring. ofc if you learn the Formula you can do it all, and that's why the school only went with that, but damn factoring simple looking quadratics are so much faster tests become a lot easier. One of the few things from basic school that I kept hardcoded in my brain. I looked at this equation and took me 4 seconds to solve it. If you like Math (or is just quick calculator), I advise everyone to learn both ways cuz factoring speeds solving up so much you end up having way more time to solve the problematic (non factoring) ones.
In Germany we usually get taught a slightly different formula with only two variables p and q, after dividing the equation by 2 (in this case).
x^2 + p x + q = 0
The 2 solutions then are:
x = -p/2 +/- sqrt((p/2)^2 - q)
EDIT (removing accidental strike-through):
x = - p/2 +/- sqrt((p/2)^2 - q)
I always find it weird seeing that much more complex formula compared to our "just divide by a first" approach 🤷♂
@@silentguy123 The formatting messed up the formula, so that was probably a bit confusing.
Actually I used that formula to work out both solutions in my head after just looking at the thumbnail. In my opinion it's a bit easier to remember than the a-b-c formula, but it may just be because that's what I was taught.
Here are the steps I did in detail:
2 x^2 + x - 6 = 0
divide by 2 to get rid of the factor in front of x^2:
x^2 + (1/2) x - 3 = 0
So p = 1/2 and q = -3.
Now substitute p and q into the formula (from my original comment) to get
x1 and x2
= - 1/2/2 +/- sqrt((1/4)^2 + 3)
= - 1/4 +/- sqrt(1/16 + 3)
= - 1/4 +/- sqrt(1/16 + 48/16)
= - 1/4 +/- sqrt(49)/sqrt(16)
= - 1/4 +/- 7/4
So x1 = - 1/4 + 7/4 = 6/4 = 3/2
and x2 = - 1/4 - 7/4 = - 8/4 = - 2
:-)
I have zero use for any of this information. I'm almost 8 years into my career of treating people with pain which requires zero math. I still found myself watching this entire video and subbing. Something about it was just captivating. Maybe I'll be able to use this information when I'm helping my kids. Thanks!
Eh? You don't deal with dosages, thresholds, and body weight proportions?
@@abominationdesolation8322 Yes, but in a more nuanced way. I don't need to run equations or math to come up with things that I use. I utilize mostly a graded exposure approach and adjust it based on symptom/movement response. I know there's underlying math in there somewhere haha, but I don't have to actively do any of it in my head or on paper.
It's good for the brain and you need your brain :P
I know you get this all the time but I really *do* wish you'd been my maths teacher in school!
Thank you for your nice comment!
I don't recall ever doing factoring and I got all the way through dif eq and calc3.
I feel cheated.
Calculus is way more useful than quadratic factorizations. I am 20 years removed from school and I still remember how to do calculus (including multivariable differential equations) and I still remember the quadratic formula but ask me to factor a quadratic now and I'm deer in headlights.
0:53 isn’t it equal to 0, not 6
Ah yes! I said it wrong in the video but luckily I wrote it down correctly 😊
love this channel, been watching some of your vids and they make me realize how much i used to enjoy math, and how much ive forgotten
So much fun! Just discovered this channel. I’ve always loved math. Now in my 60s, Ph.D. chemist.
Thank you!!
I did it in my head successfully, it’s -2, but honestly I might not be able to solve it on paper if the mental math were harder. I’m getting stumped on what you can actually do to isolate the x cleanly
I did in my head as well. The answer is 1.5
The method where you factored the thing is called "Middle term splitting" as I was taught. It is the same thing but instead focusing on the first and last term, we focus on the Middle term. In this example 2x^2 + x - 6, we first multiply the coeffecient of the first term and multiply it with the constant (2 and -6) to get -12. Then we have to find two coffecients of x so that when they are added, they give x as the sum, and also multiply to give -12 as product. In this case, they are (+3 and -4) and (-3 and +4). Then we split the middle term (x) in this way (2x^2 -3x +4x + 6) and then the equation is solved further, (2x^2 - 3x)(4x + 6) = 0, which will be solved to get both answers as you showed.
But (2x^2 - 3x)(4x + 6) = 0 is not the same as (2x^2 - 3x + 4x + 6) = 0. The first results in 8x^3 - 12x^2 + 12x^2 - 18x = 0 => 8x^3 -18x = 0. This results in answers of x = 0, x = 3/2, x = -3/2.
@@malindemunich2883 My mistake lol, it will actually be (2x^2 - 3x) + (4x + 6). It is a factoring trick that I was taught
He forgot to factor out the x, but he remembered the jist of the method correctly
@@richfutrell753 - Yeah, I wasn't trying to be hostile. I honestly hadn't seen that method before and was trying to see what would be the correct next step, so the idea is to eliminate the additional x from the original factor set? That seems reasonable.
In other words, (just to clarify, because I am interested), his initial result of two factor sets would be correct excepting that you don't include the x on the second term of the first factor, not his correction of using the addition sign between them, yes? Because that seems to then create the correct solution.
@@malindemunich2883 Even the greats make mistakes. I will look out for any mistakes in the future
I sort of like throwing away the quadratic formula and always complete the square.
So we aim for x^2 + 2 B x = C, here B=1/4 and C=3.
Then we add B^2 to both sides.
x^2 +2Bx +B^2 = C+B^2.
The left is just (x+B)^2
(x+B)^2=C+B^2
(X+1/4)^2=3+1/16=49/16
Taking the square root:
(X+1/4)= +/- sqrt(49/16)
X+1/4= +/-7/4
X=3/2 or -2
(X-3/2)(X+2)
we divided by 2 originally, so throw that in:
(2X-3)(X+2)
The quadratic formula is just a memorized version of that. But it isn't much faster than doing it by hand.
Here you don't have to resort to the quadratic equation. There is an interesting way to solve this that my teacher taught me a while ago. I'll write the steps below.
Subtract 6 from both sides.
2x^2+x-6=0
Multiply the leading coefficient (2) with -6.
x^2+x-12=0
Factor using the factors -3 and 4.
(x-3)(x+4)=0
Divide the constants by the 2 that we originally multiplied them with.
(x-1.5)(x+2)=0
Set each part equal to 0 and solve.
x = 1.5 (3/2), x = -2
You could also do the method where you multiply a and c to get 12, see what mumbers add to +1 and multiply to 12 (-3 and 4) then substitute them jn for +1 to get 2x²+4x-3x-6. Split them apart (mentally) to get (2x²+4x)(3x-6) factor to get 2x(x+2)-3(x+2) to get the final answer (2x-3)(x+2). Then it's as simple as mental addition.
Judging by the picture i thought this was supposed to be a trick question because if you flip the image upside down, it looks like e = x + e^xe 😅
That one is also easy to solve
@@munteanucatalin9833 I know about Euler's number and its value but idk if it has some special properties or something when it occurs in exponents. So I'm not sure how to solve it yet.
@@dynaspinner64 That equation is a basic type of Omega function called Lambert W function. Just rewrite the equation using Lambert formula: 1=(e-x)e^(-xe) and make the appropriate substitutions then simplify it and solve for x as a normal exponential equation. Haven't you studied in college beta, Delta, gamma, sigma Omega functions? This is second term Math for a BA in Physics.
@@munteanucatalin9833 I'm a 11th grade high school kid.
@@dynaspinner64 O.ooo... Then keep studying. Math is a nightmare, but is useful and has a lot of practical applications.
Here's how Indian students do it:
2x² + x = 6
(Bringing 6 from R.H.S to L.H.S)
2x² + x - 6 = 0
(Splitting middle term)
2x² + 4x - 3x - 6 = 0
(Taking 2x as common in the first and second term and 3 as common in the third and fourth term)
(x + 2)(2x - 3) = 0
Hence, x = -2 (or) x = 3/2
I’ve started watching your videos to kinda test myself on math equations I used to know how to do. It feels good that I can keep up and still learn/relearn these lessons. Thank you.
The method of middle term splitting is very common in india and we have vedic method of factoring all this is a class 7 maths in india
India wale like kare
Mujhe nahi pata tha ki middle term ko split karna vedic method hai😮
Yes it is a vedic method
I am very happy with myself for finding the solutions in my head. First I tried +2, and got 8=6, so I saw that -2 would work by guessing. Then to get the other I used the quadratic formula (to a degree since I knew one solution I didn't need to do full version) and got 1.5
Same, took me 10 seconds and I am not a genius ...
I only found the 1.5 one, I never considered using negative numbers.
Solve by completing the square!
2x^2+x=6
clear the leading coefficient: /2
x^2+1/2x=3
make a perfect square by adding half the middle term squared to both sides: + 1/16
x^2+1/2x+1/16=49/16
left side is now a perfect square as follows
(x+1/4)^2=49/16
clear the ^2: square root
x+1/4=±7/4
move the 1/4
x=-1/4±7/4
clean it up
x=-2 or 3/2
Fun times!
If only this channel exist during my school years. I cant digest this during that time, but looking at this now it make sense. 😢
That person from the post in the beginning probably forgot to subtract 6 on both sides... That would be the most common mistake anywa
Students can get so frustrated when they get to this level of algebra. They might be motivated to push through if they knew that this is basically the hardest thing to do in math until they get to differential equations. Get a solid understanding of advanced algebra and calculus is a breeze.
I love your channel because it gives me a nice review of things i haven't done since high school or college.
I had never seen the tic tac toe box method but i did know it had to make a middle from factoring.
This was an interesting method. Thanks
Even for someone who is already finished college, this is still a refreshing thing to remember. Stupidity is when you don't want to learn, not when you are unable to learn/solve a question
2x²+x=6
2x²+x-6=0
2x²+4x-3x-6=0 [middle term split]
2x(x+2)-3(x+2)=0
(2x-3)(x+2)=0
So either of the two must be zero because their product has to be zero
if 2x-3=0, x=3/2
if x+2=0, x=-2
Easy peasy
If you want to know how middle term split works, here I am
eqn form is ax²+bx+c=0
Find a*c
Here a=2 and b=6
ac=12
Now prime factors of 12 are the lowest value prime numbers you need to multiply to get 12
12=2*2*3
Now make two sets that on addition or subtraction give 'b'
If the last value is negative you need to do subtraction and if positive addition
Here the two sets can be
4x-3x
Now take something common from two parts of the eqn and solve it
Using the same, solve
Easy: x² + 2x + 1 = 0
Hard: x² + 2x - 3 = 0
Equation is (+c), so how come (-6) is utilised
Both are easy
@@himalayavalvi4651 value of c is -6
So bx+(-6)
bx-6
@@aeojoe clickbait it is 😉
From the beginning just divide 2x²+x-6=0 by 2 to get x²+0.5x-3=0
Then figure out what numbers add to 0.5 and multiply to -3.
2 and -1.5
(X+2)(x-1.5)=0
X=-2 and x=1.5
@AmmoGus1: You are also correct and you also prove that it does not work, the guy in the video also prove it does not work, with both his method, because the rules of algebra x must be equal to only one value, not 2 diffrent value,. Lets plug in your answer to found out if both answer work 2(x^2)+X=6 lets replace all X with -2= 2[(-2)^2]+2=6 2(4)+2=6 8+2=6, 10=6. That mean x=-2 is wrong. 2[(1.5)^2]+1.5=6 2(2.25)+1.5=6 4.5+1.5= 6 6=6 So the 1.5 actually work. It does not work because you found one incorrect answer and one correct answer. This is why it does not work. This is the solution replace X^2 to Y^2 so that way your formula as no problem because it would go like this Y=-2 X=1.5 and now solve it 2[(-2)^2]+1.5=6 2(4)+1.5=6 8+1.5=6 you see even this does not work. So no matter what it is always undertimine. If they is always more then one answer that work, or does not work, it is undertimine.
@@SuperNickidthat isn't true at all, the equation x⁴-5x²+4=0 has 4 solutions: 1, -1, 2, and -2. All of them work, you can try it yourself. X is not limited to one value. In fact, the way I came up with that equation was by multiplying 2 quadratics with that each have 2 solutions, x²-1=0 and x²-4=0. The first one's solutions are x={1, -1} and the second's solutions are x={2, -2}.
@@SuperNickid Can you explain to me how exatly you managed to make x=-2 wrong? 2*-2^2+(-2)=6. 8-2=6.
If you replace X with -2 it = 2[(-2)^2]-2=6 and not 2[(-2)^2]+2=6. There for 2(4)-2=6, 8-2=6, 6=6. In a quadrtic equation W will always have 2 answers and you can graph the equation and see the two points where the graph crosses the X axis@@SuperNickid
@@SuperNickid You substituted wrong in the x=-2 case. You wrote "2[(-2)^2]+2=6". You correctly substituted x with "-2" on the first term but you incorrectly substituted x with "+2" on the second term. You should have written "2[(-2)^2]-2=6", which simplifies to "6=6" and checks out.
For some reason people hate completing the square but I find it easier than factoring.
Standard form for reference: ax²+bx+c
2x²+x=6
First, make sure that your c (in this case 6) is on the right hand side.
Then, divide everything by a
In this case you have x²+0.5x=3
Then divide b/a (in this case 0.5) by 2, then square it
You then have (0.25)² as a result
Now add that on both sides
x²+0.5x+(0.25)²=3+(0.25)²
Notice that you have a polynomial on the left hand side that is a perfect square
You can now condense it to (x+0.25)²
So now you have (x+0.25)²=3+0.0625
Simplify to (x+0.25)²=3.0625
Then you take the square root
So we have x+0.25=±√(3.0625)
It would be easier to use fractions in this case because you can distribute the square root
In any case then we have x+0.25=±1.75
So we have x=±1.75-0.25
Split this into x=1.75-0.25 and x=-1.75-0.25
And then you have x=1.5 and x=-2
Note that completing the square is the proof to the quadratic formula (Use the base equation ax²+bx+c and do the exact same steps as above)
Anyways you probably have work on completing the square that I didn't find and I like your videos :)
The quadratic formula lives rent free in my head thanks to my teacher. To the tune of "pop goes the weasel":
X equals the opposite of b, plus or minus square root, b squared minus 4 a c, all over 2 a!
It’s a simple quadratic equation, solved by x = -b yada yada yada. 😎
finally my practice blessed me, as soon as I saw the problem I figured it out
I feel good
especially because I always got bad grades in my tests
I feel confident
Sorry for bad english it's because I don't respect the language
btw a random help, if you can please
I am a high schooler, I especially like math can understand concepts and am having fun with it
though still I get bad grades
any suggestions
Well, it really depends on how much you practice. Understanding concepts and solving questions are two different things. You could understand the concept, but not understand the question. Sometimes, they twist the question to make it harder for you to understand. Or... you take a lot of time to solve simple questions. That again, can be solved by practicing questions by limiting yourself to a certain time period.
@@cacalightx that's the fun part
I don't know; I practice hard, I do everything I can possibly do but still the same low result
And the main thing it's always the exams or tests that I cannot do math, whenever I see those questions on the exam(same difficulty) I cannot, I also try to do those same problems given in exam/tests at my home, and guess what I can do them easily
I don't know what's the problem with me
Sorry if it's a bit confusing, English is not my native language so I couldn't explain properly
@@kafkatotheworldcould be your teacher's fault. Some teachers are horrible at explaining things. Pay all the attention you can in class,do all your homework and you should be golden, but there's not much more you can do if your teacher sucks. Ask questions if you don't understand something.
@@kafkatotheworld maybe its psychological and the exam environment hinders you in some way. Are you too stressed? Are you too anxious? Does it distract you? Take a deep breath and ground yourself.
Bro who practices math
I have been tutoring and teaching algebra for nearly twenty years and a random guy on RUclips teaches me a way to show how to figure out factor terms in a quadratic that is so much easier.
I love that you did a simple problem that's still tricky to those of us that have been in some advanced math for a while
My brain at first sight : -2
The thumbnail: Uncle Chow
He’s like Uncle Sam, but instead of trying to get you to draft, he’s trying to get you to math.
I really appreciate that you showed how the first method of factoring can still work with coefficients on the x^2. Whenever I saw that in my math class I got so confused because my teachers never showed us how to factor with coefficients like that, they just immediately pointed to the quadratic formula.
I always struggled with the factorising method when the coefficient of x²≠1. I never knew that method that you used, even when I was a maths teacher.
I learned the quadratic formula in either 7th or 8th grade to the tune of jingle bells and will never forget it.
I never knew about such a clear and straightforward method for factoring when I was learning this. This tic tac toe method would have been amazing!
U can use Shredharacharya Formula to solve this x=-b +- √b²-4ac upon 2a ...
We did these type of questions in class 4
When I was at school we used to (and had to, btw) actually calculate it instead of using the formula. So several decades later I tried to do so and could still do it:
2x² + x =6 | Divide by 2
x² + 1/2x = 3 | complete the square i.e. add 1/16 *explanation below
x² + 1/2x +1/16 = 3 +1/16 | apply 1.binomial
(x +1/4)² = 49/16 | extract root
| x + 1/4 | = 7/4
x + 1/4 = 7/4 or x + 1/4 = -7/4 | -1/4
x = 6/4=3/2 or x = -7/4 -1/4 = -8/4 = -2
*completing the square: We want to reduce this equation using the first binomial formula, i.e. (a+b)² = a² + 2ab + b². We already made a equal x, so we only need to calculate b. With 2ab = 1/2x and a = x we get 2b = 1/2, so b = 1/4. Now we only need b² which is (1/4)² = 1/16. We add this to both sides of the equation and can consequently apply the first binomial formula
Excellent explanation and demonstration.
we weren’t allowed to use the quadratic formula (or PQ-formel in Swedish, as it is a simplified version of the regular one) in our first uni math course for some reason I can’t remember, instead they wanted us to finish the square* even though they both yield the same result. I think the reasoning behind the rule is because the quadratic formula comes from finishing the squares
*if you do not know what finishing the square is, I’ll try my best to explain it below:
suppose you have ax^2+bx+c=0. then you factor out a so you get a(x^2+(b/a)x+c/a) = 0. then you look at the coefficient for x. when expanding a binomial (α+β)^2 you get (α)^2 + 2αβ + β^2, so to ”reverse” operation that you would take the α square and the 2αβ and divide both by α, and the remainder of 2αβ by 2 since the expansion is αβ + αβ. that would leave you with (α+β)^2. doing the same ”reverse” with the original question i posted, you would ”divide” the x squared part and the (b/a)x part with x, and also divide b/a with 2, leaving you with a((x+b/(2a))^2)+c/a)=0. the thing is though, we can’t say the equations are equivalent with one another since expanding the square in the second form would produce an extra +(b/(2a))^2, so in order to have equivalency we must subtract this outside the brackets, so we end up with a((x+b/(2a))^2-(b/(2a))^2+c/a) = 0. from this point you often remove the a for simpler calculations (but remember to put it back in the end) and you move forwards with putting the terms outside the squared brackets on the right side, so we would get (x+b/(2a))^2 = (b/(2a))^2 - c/a
squarerooting both sides gives
x+b/(2a) = +-sqrt((b/(2a))^2-c/a) x = -b/(2a) +- sqrt((b/(2a))^2-c/a)
from this you can factor out 2a as a denominator for both terms in the right side, and to factor in c/a we find so sqrt(c/a) = sqrt(H)/2a => c/a = H/4a^2 (c/a)*4a^2 = H = 4ac. Finally, putting all this together we get the final expression of X = (-b+-sqrt(b^2-4ac))/2a, which is the quadratic formula
anyways, the point is that because of this rule i’ve grown fond to use this method instead even if it is longer and if we’re allowed to use quadratic formula during exams
@@martineriksson03 The derivation you give of the ABC Formula (as it is known in Sweden and some other countries, but not in English speaking countries) relies on _completing the square_ as it is known in English. In my language this is known as _splitting off the square_ which is not quite the same but the two terms are mostly used interchangeably.
While solving quadratic equations by completing the square or to derive the ABC Formula you can avoid the use of fractions until the very end using what is currently known as Sridhara's method but which was referred to in high school algebra books throughout the 19th century as the Hindoo method. This is by far the easiest method to derive and prove the quadratic formula but can be somewhat less practical in solving actual quadratic equations by completing the square because the coefficients tend to become large. Let's start from
ax² + bx + c = 0
where a, b, c are assumed to be real numbers and a is not zero, otherwise we would not have a quadratic equation. First subtract c from both sides which gives
ax² + bx = −c
Now multiply both sides by 4a to get
4a²x² + 4abx = −4ac
or
(2ax)² + 2·(2ax)·b = −4ac
Add b² to both sides to make the left hand side a perfect square and we have
(2ax)² + 2·(2ax)·b + b² = b² − 4ac
which can be written as
(2ax + b)² = b² − 4ac
Assuming b² − 4ac is not negative this gives
2ax + b = ±√(b² − 4ac)
Subtract b from both sides and then divide both sides by 2a to get
x = (−b ± √(b² − 4ac))/2a
and we are done.
This method is attributed to Śrīdhara also known as Śrīdharācāryya or Śrīdhara Acharya but commonly spelled Sridhara in English sources. Sridhara was an Indian mathematician who lived in the latter part of the 9th and the early part of the 10th century. His original text where he discusses his method appears to be lost, and we only know about his rule from a quotation in the works of a later Indian mathematician, Bhāskara II, who lived from 1114 to 1185.
In India the quadratic formula is known as the Śrīdharācāryya formula and in Brazil the quadratic formula is known as Bhaskara's formula. In parts of Germany and Austria it is known as the Mitternachtsformel (midnight formula) because, as the story goes, students were required to remember it by heart even if they were woken up in the middle of the night.
Going through school and actually learning how to do these problems that are considered tricky without needing the explanation is the biggest flex (I’ll never use this again).
It looks less scary when you do (Greek letter delta here) = b^2 - 4ac, and then x1 = -b-sqrt(delta)/2a and x2 = -b+sqrt(delta)/2a.
Bruh, I used to solve these in seconds when I was in grade 8, now that I have done masters in biology, this looks the hardest thing, when I can't remember how to solve it. 😭😭
Love the dose of nostalgia when listening to your detailed explanation of each step.
I equally hate all those teachers who just throw the quadratic equation to their students without explaining what is it, why is it and without introducing the concept of discriminant.
The formula is just a big lump of algebra without those explanations.
the discriminant which I usually call Δ, is the bit under the square root. Meaning Δ=b²-4ac. It is so useful as it tells you how many real root you should expect.
If Δ＞0, expect 2 real root, thanks to the ± in front of the square root.
If Δ=0, expect only one real root
and if Δ＜0, expect no real root.
This serve as a really good gatekeeper for your answer.
and pkus, the formula is now as short as (-b±√Δ)/2. A bit easier to remember now isn't it.
I started factoring by using the graph. It's so much faster than guess-and-check.
Thank you. This is mostly a review for me, but it’s very helpful. So thanks!
All the videos of him I've seen, I'm still impressed with his marker control.
Get that six to the left side so you get zero on the right so you get a x squared + bx + c equals zero shape. Then, use the formula for quadratic equation and problem solved.
I haven’t seen that chart method for cross multiplication before, but it feels like a good way to keep it organized and reduce the amount of mental math necessary
I was never very good at factoring or completing the square...but I quickly memorized the quadratic formula and thats how I always solved quadratics in school.
I never memorized the quadratic formula, and I never liked cross-checking when factoring, so I did it the hard way: Completing the square.
2x^2 + x = 6
x^2 + (1/2)x = 3
x^2 + (1/2)x + 1/16 = 3 + (1/16)
x^2 + (1/2)x + 1/16 = 48/16 + 1/16
x^2 + (1/2)x + 1/16 = 49/16
(x+1/4)^2 = 49/16
x+1/4 = +/-(sqrt(49/16))
x+1/4 = +/- 7/4
x= +/- (7/4) -1/4
x= (-1 +/- 7)/4
x= -8/4 | 6/4
x= -2 | 3/2
I wish I’d had someone like you as a math teacher or at least videos like this when I was in school. I felt stupid A LOT in math class, often ended up I in tears doing my homework or because of a bad grade on a test, and still sometimes have bad dreams about impossible math. (Somehow though, despite all that, I wound up in a job where I do basic data analytics. But I think it’s a lot easier for me to handle statistics and data analytics because it’s computer assisted and because I understand the type of data and real world applications I’m using it for and it’s not happening in a vacuum like it does in school.)
just multiply the coefficient of x² and the constant. factor out the product of those two such that you get two factors which account for the middle x term of the equation.
when this doesn't work, i use the quadratic formula
Quadratic equation is like the bread and butter of algebraic equations in my middle school math class.
The first solution seemed to need more trial and error than is strictly necessary. We need to find two numbers which add to b and multiply to ac, in this case adding to 1 and multiplying to -12, which (as -12 is negative) requires two positive numbers which multiply to 12 and have a difference of 1. Clearly the positive numbers are 3 and 4 and we adjust these to get -3 and 4 as the original numbers. So the equation is (2x+4)(2x-3) = 0 which can easily be solved. You start off with an "a" in both terms but it should cancel away, in this case you can write (x+2)(2x-3) = 0. In more complicated cases part of the a will cancel in one term and the rest in the other.
Fast method: multiply A times C. Find the factors of that number that would add up to your b value (in this case 2×-6=-12 so youd factor -12 into -3 and 4 which -3+4=1) from there you would divide them by the a vaule if they can ans then put it into factored/intercept form. (In this case it is (x+2)(2x-3)), then set one of them = to 0 an solve for x. Then do the same for the other. (X+2=0 x=-2, 2X-3=0 x=3/2) This method WILL ALWAYS work IF the factors can add up to the B value. If not, use this or quadratic formula (i'd use the formula).
Really great teaching! Thank you. 😀
I studied in my school but still dont know where these equations are used. I think its important to study why need to learn it rather than solving equations. After 12 years I realized the importance of calculus. In my school I was just solving it for exams
What would a practical example be in the working world of manually calculating this equation? (not relying on calculators, computers and such).
Help me out here.
Since you can rely on computers, there are very few practical reasons why you have to learn how to do this manually. Any computer with an internet connection can access Wolfram Alpha, which will solve it for you. The main reason to learn this, is so you understand what a computer is doing behind the scenes to solve these problems, so you can give a reality check on the answer.
One reason where you would want to know how to do algebra manually, is that sometimes algebra problems you may develop for a practical application, will be too computationally intensive for the computer to handle. Knowing how to preprocess it, with manual algebra, can greatly simplify the work the computer has to do.
As an example, consider this system of equations I was trying to solve for a practical application. In this application, we want to know x and y that satisfy these three equations. We don't care to solve for L; as long as it is consistent between both equations. If you are curious about the application, look up Lagrange multipliers and constrained optimization.
4*x^2 + 9*y^2 = 9
-1/2*x*(9*x^2 + 4*y^2 - 8)*e^(-x^2 - y^2) = 8*L*x
-1/2*y*(9*x^2 + 4*y^2 - 3)*e^(-x^2 - y^2) = 18*L*y
This problem stumped Wolfram Alpha, but with a little preprocessing, I was able to figure out how to simplify Wolfram Alpha's job, and make it much easier to get an answer.
The second two equations show us that either x=0, or y=0, will make a trivial solution. So it's as simple as finding the other corresponding value of the other variable, from the first equation. Once we rule out the trivial solutions, x will cancel out of the middle equation, and y cancels out of the final equation. We can also shuffle the -1/2 to the other side, by multiplying by -2.
At this point, we notice that all instances of x and y are both squared. So we can assign a temporary variable to represent each of them. Let V = x^2, and let W = y^2, and this gives us:
4*V + 9*W = 9
(9*V + 4*W - 8)*e^(-V - W) = -16*L
(9*V + 4*W - 3)*e^(-V - W) = -36*L
This is now in a form that Wolfram Alpha can solve. The solutions for V & W are:
V = 72/65, and W = 33/65
Which implies that the non-trivial solutions for x & y are:
x = +/- 6*sqrt(2/65)
y = +/- sqrt(33/65) of the same sign as the corresponding x
The trivial solutions are:
x=0, y = +/- 1
x = +/- 3/2, y = 0
One place I've used it, or would've used it if I were doing my job prior to computers, is to calculate the ideal points to support a beam.
Consider a 10 ft beam carrying a uniform load, that will be supported at two locations along its length. Determine the two locations that maximize the load it can carry.
At the end of this problem, you'll end up solving the following quadratic equation:
1/2*x^2 + x*L/2 - L^2/8 = 0
Where L is the total length of the beam, and x is the position of the first support. L-x is the position of the other support. The solution ends up being x= L/2*(sqrt(2) - 1), which equals 2.07 ft. This means you'd have a span of about 6 ft, and equal cantilevers of about 2 ft, for a 10 ft beam.
🤔 I haven't seen the box technique. I divided everything by 2 and asked what adds to positive 0.5x and multiplies to negative -3, and got (x+2)(x-1.5)
I only saw the thumbmail and needed like 5s to solve X= -2
I never thought that a math video would be so comforting. Thanks for the detailed explanation, man!
I learned the formula 50 years ago and it is stuck in my head to this day! 😂
To factorise, I've always been taught that-
ax²+bx+c
Separate bx into such a manner that the product of their multiplication is equal to ac and the sum of them equals to b.
So 2x²+x-6 would be
2x²+4x-3x-6
2x(x+2) -3(x+2)
(x+2)(2x-3)
I wish youtube math explanations were a thing back when I was in school.
Here I am reviewing this stuff late at night because I forgot all of this. Gotta keep my brain sharp! I really appreciate your clear explanations.
Simpler method for those who have difficulty going directly from the quadratic to the factors. For ax^2 + bx + c = 0, calculate a*c, here 2 * -6 = -12. Now look for 2 numbers that multiply together to give this number and add together to give b. Those numbers here are 4 and -3. Now replace b with these 2 numbers: 2x^2 + 4x - 3x - 6 = 0. Now group terms 1 and 2, and terms 3 and 4, so that they have a common factor: 2x(x+2) - 3(x+2) = 0. Combining: (2x - 3)(x + 2) = 0. It is a little longer than the methods most schools teach but it each step is simple meaning that you will never go wrong, whereas I see errors in the direct factorisation all the time. Any school pupils who have trouble with this process, please try this!
really cool video. when i was in grade it i was taught to solve this some other way and we dint have quadratic formula until 10th grade, we used to split the middle term and then take the common factor .
6:30 Any reason why not using the pq-formular? x1/2 = -p/2 ± sqrt (p^2/4 - q) Honestly, I don't really understand why people (and books) use the formular with abc - if you can simply devide by a.
its fun to watch these videos.I havnt done this type of math in years. Its a nice refresher.
I don't blame the eighth grader. I blame the teachers who don't teach their classroom students basic algebraic concepts. They are more interested teaching social engineering than mathematics.
I haven't solved any quadratic equation since school or university, but I still remember quadratic formula. But here is CIS we got it in another way. We used to find D (b^-4ac) and only then find x1 and x2
X = 1.5 or -2 you can just do it in your head.
If X = 1 then it's 3
If X = 2 then it's 10 so the answer is in-between those, X = 1.5 is 6 by approachment.
Since it's to the power of a even number we need to check if there is a negative answer, X = - 1 = 3 and X = -2 is 6 so -2 is the other answer.
When we put in 2 for X we could have guessed that the negative is also correct because the difference between the positive and negative number in the answer is always double the number put in to X (because you subtract X not add it since X is negative) so if X= 2 is 10 then by subtracting 2 times X you get the negative answer which is 6
X = 3 would be 21 so by my logic X = -3 should be 15 and calculating it we get 15 as the answer.
I solved it by immediately seeing -2 as a solution, then finding the minimum of the parabola by factoring out (a^2 + 2ab + b^2) from the formula. Found it at -1/4 and mirrored -2 on it to receive 1.5.