Awesome Math Question

[@Tensto]

I love the interpretation of the negative result, really puts things into perspective

[@happybuggy1582]

What perspective

[@massimookissed1023]

Yeah, it shows how there can be two different x values that both make sense, and why one of them is ignored, rather than just arbitrarily saying ignore the -ve value derived from the equations.

[@Duke_of_Lorraine]

Just because a result is mathematically correct doesn't mean it's true, indeed

[@Patrik6920]

@@Duke_of_Lorraine ..Jake just proved both roots r true, but ya we usually diregard the roots that doesent make sense... how ever, we can insribe this into a coordinate sytem and have the square are atart at the coordinates [(5 +5√17)/2, (5 +5√17)/2 ], wich leads to the side of the square being ((-5 ±5√17)/2)² = (10 * 20 - ( (-5 ±5√17)/2 )/2 OR ((-5 +5√17)/2)² = (10 * 20 - ( (-5 +5√17)/2 )/2, ((-5 -5√17)/2)² = (10 * 20 - ( (-5 -5√17)/2 )/2

[@rosettagrey2851]

Negative solution interpretation blew my mind, the placement changing based on negative/positive options and the areas still being equal was very exciting indeed.

[@keijzer83]

You were really thinking outside the box

[@aaditnag7960]

literally 🤣🤣

[@jamesrocket5616]

Literally

[@JakeArnet]

Underrated comment.

[@Flyore]

Negative Solution blew my mind , loved it !

[@frankstrawnation]

The interpretation of the negative result was great! Please, do it more times.

[@Insightfill]

I remember in trajectory problems in physics we'd explore the negative result of the quadratic answer. Also: sometimes collision problems would give two positive answers and you'd use the smaller positive one, as the larger positive answer would represent the SECOND collision.

[@Jerom_]

Collision problems should always give you two results: one result of the actual collision, and one result where the bodies do not actually collide but fly through eachother. That is because for a collision problem, we solve for conservation of momentum and conservation of energy. And the momentums and energies not changing clearly meets this requirement.

[@Insightfill]

@@Jerom_ Yes, it was always entertaining in those pre-Internet days to imagine a car getting rear-ended and then the other car drifting through it - out of phase - and then the rear-ended car "gets even" and hits him back!

[@m3m3sis]

that negative extra in the end was amazing, more of these please! Extremely insightful!

[@aaditnag7960]

Andy I recently started watching your videos and I have never really been attached to only one content creator - but you have changed that. Your method of teaching is so simplistic and clear - I hope you never stop making these fun, interesting and amazing videos. Keep it up - you've earned yourself a like and a subscriber. How exciting!

[@Epyxoid]

That's how it goes! Negative answers sometimes give you an out of the box solution. 😄

[@thedrdroppedme]

Hey Andy! I've just started relearning math after 30 years and found your channel 2 weeks ago. I've relearned geometry from scratch here, and I can't thank you enough. That said, I think I found an even easier solution than your video! Let me know if this makes sense: Since the blue triangle is always 1/2 b/h, the sum of the area above and below the blue triangle will ALSO be the same area! So without even calculating height of the triangle, we can reason that the Red square is X^2, the Blue triangle is also X^2, and the sum of the white triangles above and below the blue triangle are ALSO X^2! This means the entire puzzle can be calculated by solving for the missing white rectangle above the red square - which has a height of 10-x, and a width of x. So the entire equation should be able to be solved as 3x^2+x(10-x)=200! If this is correct, I'm gonna be so happy, because I'm just getting started relearning math!

[@ghostbluster8900]

You get the same quad formula except with a factor of 2. Great thinking

[@kurtlindner]

Taking that extra step to fully explain the negative solution is why your better than the 5 best math teachers I had in school combined.

[@jpg7616]

1:25 why would you subtract the x^2 and then inverse? Wouldn’t it be easier to just move everything over to the x^2 side? Add 5x to both and subtract 100 from both.

[@aaronross2578]

Yeah, you're right. Both methods are right. It's probably how he learnt it or even better he is explaining everything step by step so that someone who doesn't know algebra can understand too

[@loancell]

Love these math questions! Keep it up!

[@nmklpkjlftmch]

I did it your way and then saw a different way that I found visually easier. Continue the line from the right edge of the red square to the top, giving a small rectangle above the red square that's (10-x)*x and another rectangle containing the blue triangle. You have the red square and the blue triangle which are both x^2. The blue triangle is half the area of the rectangle it's inside, so the rectangle containing the blue triangle is 2x^2. The whole rectangle is 20*10 200 = 3x^2 + (10-x)*x = 2x^2 + 10x Subtract 200 from both sides and divide both sides by 2 to get x^2 +5x - 100 = 0

[@dorchevsky2556]

Truly fascinating when you decided to explore the negative solution

[@yamazakuraaaa]

I loved your solution, especially the negative part. Never thought that could happen. I ended up solving this question in another way leading to a slightly different answer. I could be wrong but my solution is merely based on high school level, so please excuse my mistakes. so the first step was to find out the area of uncolored parts (the trapezium and the right triangle) -> deduct it from the area of the rectangle -> find the sum of red and blue areas (which I wrote as twice of red area) -> put it in the equation (the below suggestion will make this part clearer) -> and solve for x. Area of rectangle = 20*10 = 200 sq units Area of trapezium (please notice the right angle) = 1/2*base*height = 1/2*(20)*(10-x) = 100-10x sq units Area of right triangle = 1/2*base*height = 1/2*(20-x)*(x) = 1/2*(20x - x^2) sq units Area of square (red area)= side*side = x^2 sq units RED AREA = x^2 = BLUE AREA (Given) RED AREA + BLUE AREA = AREA OF RECTANGLE - AREA OF TRAPEZIUM - AREA OF RIGHT TRIANGLE x^2 + BLUE AREA = 200 - (100-10x) - (1/2*(20x - x^2)) 2x^2 = 100 + 1/2(3x^2) - 10x 4x^2 = 200 + 3x^2 - 20x (Multiplying 2 on both sides) x^2 + 20x - 200 = 0 solving for x, x = [- 20 ± (400 + 800)½] ÷ 2 x = - 10 ± 10(3)½ x = - 10 + 10(3)½ (as length cannot be negative) x = 10{3½ - 1} x =10{1.732 - 1} x = 10*0.732 x = 7.32 units (solved)

[@TheWorldsLargestOven]

I'm actually proud that I somehow got this right... They haven't even taught the rest of my class the quadratic formula.

[@someonespadre]

I’ve plotted the other solution for fun too. Sometimes if there are 2 variables the 2 solutions can fit either one. In other words you get two correct answers.

[@rcleme]

Holy shit dude! The negative answer explanation BLEW MY MIND. You're the freaking best

[@henrygoogle4949]

Wow that 3, 2, 1 countdown got serious!

[@fuglbird]

Finally a problem where you found the simplest solution. I'm glad to see that you are learning. Keep it up a couple of years and you'll be able to solve some interesting problems.

[@mr.d8747]

*That's really thinking outside the box*

[@stajsud4618]

I did this in a WAY different way, but got the same answer! Instead of solving for the height of the blue triangle I took the areas of every section and added them, setting them equal to the total area of 200

[@vanillacube8159]

Your channel really helped me jog my memory during summer

[@d.on.in.a]

I found the interpretation of the negative solution very exciting! ❤

[@ZdenalAsdf]

The fact that we often discard negative solutions like that makes me sad. There's a whole world of solutions to problems that you're just throwing away. I guess the argument is that "you can't have a negative length" or something like that. But if you imagine that instead of finding the length of something, you're finding the coordinates of points (so here the bottom left of the large rectangle would be (0, 0) and you're trying to find point (x, x)), then it makes complete sense to also consider the negative solutions.

[@MikeSimoneLV]

I'm so glad I found this channel! I love your videos!

[@matttyner8804]

Wish I had this to watch when I was in school! Thanks for keeping me sharp

[@rathersane]

That demonstration of the negative solution was quite a *stretch!* Okay… I’ll see myself out now.

[@Mike__B]

I would have discarded the negative answer because a physical quantity like length simply can not be negative. It seemed like what you were doing with the negative solution is you were saying that the lower left corner is the origin and x represents the x & y coordinates in that space, but that's alright I like that interpretation as well. That said this problem was a good example of "read everything that is given to you" as to why I couldn't do it, I was so hyper focused on the shapes and the formula for various ones, thinking "ok there's a right triangle there in the negative space, and ..." that I completely didn't see that both shapes are the same area.

[@gauravvj1090]

Please explain 1:43 how can you just decide to divide by -1 ( if it's a stupid question then let it be, I wanna know)

[@z000ey]

You can always use multiplication (as well as division) when you do it to both sides of the equation, as both change the same and the equation stays. Imagine 5=5 and you multiply by 2 both sides, you get 10=10. Same goes for -1: 5=5 becomes -5=-5. The best is when you've got zero on one side, cause it stays zero, so that is why he moved all to one side of the equality sign, and also it means that your equality with variables must equal to zero, and from that you can deduce the variable. Beware on the degree of the equality though: in a linear equation (where you've got x without any potential) you'll always get the exact result. In a quadratic (where x is squared) you can get 2 results.

[@Grizzly01-vr4pn]

@@z000ey Of course, at 1:21 he could've just added 5x to each side, then subtracted 100 from each side, and he'd have ended up with the same quadratic.

[@greendruid33]

The negative proof was neat to see!

[@judekirkcruz7177]

WOW thanks for including the negative perspective solution

[@M3GH0]

I am so stupid. I found the white areas first and then I subtracted them from 200 to get 2(x^2) then I divided them by 2 and used the quadratic formula. It took so long.

[@AlexanderNash]

You're not solving for the area 💀

[@M3GH0]

@@AlexanderNash I found what is equal to x^2 then I squarerooted both sides

[@karolissad.4270]

Lol did u use (20-x) for the bottom white triagle?

[@M3GH0]

@@karolissad.4270 yes isn't white triangle x(20-x)/2

[@karolissad.4270]

@@M3GH0 yes it is. Just funny that you were that close to just directly getting the answer

[@clemensnader3865]

Which software does he use to demonstrate the different areas possible? (dragging the square)

[@wesleydellinger4344]

the math goat strikes again. How exciting

[@billycox475]

I always wondered if negative distances might exist in an infinite universe

[@massimookissed1023]

A negative distance is just a positive distance in the opposite direction.

[@djsyntic]

But now I'm curious, with the solution with x having an negative value, you have red an blue overlapping (let's call that purple). Is there a way to find the purple area?

[@FlyMiCat_yup]

I was thought that the formula for a triangle was (BxH)/2, and correct me if I’m wrong but don’t you have to divide all the parts of the equation? Wouldn’t it be incorrect to only divide b? Please explain, thank you! Also sorry I’m not English. Edit: after watching the video I see that I still have a lot to learn, but I’m still confused about what I asked earlier

[@karcha5]

What software does Andy you use for making these videos? He edits them as he speaks, effortlessly!!

[@JudithOpdebeeck]

turns out if you take this exact same problem, except make the big rectangle a square with side 10, the theory works out exactly the same, but the math is easier, because you actually get a square under the root

[@HeirToPendragon]

I added the area of the white trapezium, the red + white triangle as another trapezium, and then the blue triangle (which is just x squared) and set that sum equal to 200. After simplifying it becomes the same quadratic, which you solve with the formula.

[@adriancuevas91]

What software was used to visualize the negative solution? That part was amazing

[@Datnguyenburh]

i got this term: 4x^2 + 200 - 10x -x^2 + 20x - x^2 2x^2+10x+200=400 x^2 + 5x+ 100 = 200 (x+ 5/2)^2 = 425/4 x+ 5/2 = +- 5 (square root of 17) / 2 x= (-5+ 5 (square root of 17)) / 2 i use area of each part of the problem in x(kinda same strategy with u) but not use any in the triangle, as i had known that red area= blue area, so i just simply make it 2x^2 to perform the area of the square and the triangle

[@axlroc]

I'm so impressed that i could do this... Just learnt quadratics at 13 and it's crazy

[@wildfire_]

The negative representation is a bit of a bending of the rules here. There’s a difference between a negative value on a graph and negative width or area, in real life, it’s not negative width it’s just going in the opposite direction (thus is the relativity of reality). The general strategy for this problem is to use a algebraic method and if you graph it this is indeed the two possible outcomes and both of them work, but this is a negative x value on a graph, not a negative width/height on a shape.

[@claudiomarvel]

Yea, before I watched the solution, I got x^2+5x-100 and went "Oh, the answer must be using a non-integer"

[@GS-zj7cq]

I almost didn't see triangles height at first , and i imagine being a teacher about to set this up in a test , and finding out about the "other" height , oh they never solving this😂

[@Liwet.]

Would the problem work if the point where the red square and blue triangle connected in a third dimension?

[@toroddlnning6806]

what is the area of the blue triangle oversecting the red on the negative value of x?

[@alegitnolife]

Purple

[@toroddlnning6806]

correct, i think your right, when you mix blue and red, you get purple@@alegitnolife

[@Nova1006]

Yay! Another math video :)

[@totally_not_a_bot]

I got this one right! I'm happy about that 😊

[@aliadnan42]

how exciting...

[@_Unknown420_]

At 0 = -x^2 -5x + 100, Was it necessary to divide by (-1)? [1x1 = 1 mark] If so, why? [1x2.5= 2.5 mark]

[@massimookissed1023]

No, but as he said he didn't like the -ve coefficient. The quadratic formula still works though if the *_a_* term is -ve.

[@joe_duck]

Just started following first time I was able to solve just as fast 😂

[@PM9Video]

That was really satisfying

[@aaryasuparey03]

10th board exam students ⬇

[@elkanvinod7227]

Heyhh😂

[@elsabaker517]

what?

[@SanThawClause]

yes here ✋

[@Ryan50Ryan]

I had the procedure, I just forgot the composition of the quadratic formula.

[@jacobcombs1106]

The first time I solved for it when I was staring at my 5 root 17 minus 5 answer I thought, no that has to be wrong there is going to be some elegant simple solution that yields a clean answer and I solved this problem 3 different ways and got the same answer all three times and accepted it must be right. Watched the video and found out yup.

[@serk-s]

How does it come, that you do the solution of quadratic formulas so detailed. I mean you did it a lot times. Can you just write the solution and refer to the formular?

[@prodjignesh]

very interesting indeed

[@sxkjknjw2]

Well, since in every geometry problem there is the mention "not to scale" imma find a way to get the negative answer in

[@SpacialAssassin]

I managed to solve something that simple, yay, I’m not a complete failure.

[@Tmwyl]

Brilliant!

[@bucsredsoxredwings]

Well done, but substracting -x² is not the elegant way if you wanna end up with x². So +5x-100 is way more elegant.

[@adi96adi]

Finally I've solved one of these videos by just looking at the thumbnail

[@DanMusceac]

Very interesting with the negativ solution.

[@jtiumproductions7629]

Nooooooo I got so close to solving it on my own but I accidently forgot a negative sign at the VERY END :( Anyway Wow that’s really cool to see how you can interpret the negative answer

[@Jaded-Wanderer]

That was exciting.

[@Felipera_]

Negative solution? How exciting!

[@rudybertazzo]

It was indeed exciting!

[@mariolazzarini5806]

Amazing

[@nabil4389]

How exciting

[@MrFrmartin]

got same solution as you

[@vikthepro]

As a grade 4 I’m happy I understand all of this and what it means😊😊

[@mdforhad-wk1zo]

How exciting❤❤

[@Tarif2809]

How very exciting!

[@kenhaley4]

Your presentations are wonderful! What software do you use to animate your solutions?

[@ryotoiii]

The first one of these videos I actually did beforehand! Yay!

[@ParadoxHazzard]

Lots of love ❤️

[@klatis84]

coooooool negative reality check

[@GirishManjunathMusic]

ok so assigning h as height of triangle: we know h + x = 20 and x² = 5h ⅕x² + x - 20 = 0 x² + 5x - 100 = 0 x = ½(-5 ± √(25 + 400)) x = ½(-5 ± √425) x = ½(-5 ± 5√17) Geometric Convention requires x to be positive: x = (5/2)(√17 - 1)

[@cubles6825]

Hello! I just got done with a math competition and was stumped by a problem I thought would be cool to make a video on. 3 circles, all radius of 1 Arranged like this: O O O Ignore the gaps, they’re tangent with each other and touch at only 1 point. You must find the area of the small space between them all.

[@AlexanderNash]

Its the area of the triangle minus the area of the circle segments which is just half the area of one of the circles cause its' 3x60degrees=180 degrees.

[@Grizzly01-vr4pn]

Area = √3 - π/2 ≈ 0.161 units²

[@meefmeef6254]

1:23 why did you subtract the x^2 instead of the others there instead of the other way around you just made it harder for your self anyway this was a cool problem HOW EXCITING

[@ImMUSLIM_999]

Ppz anyone tell me why did he use 1/2 B×P Bcz this is not a rigt angled triangle😢

[@massimookissed1023]

½ base × height. Think of the rectangle 10 side as the base, then the triangle's height is from that 10 side to the corner of the red square at 20-x.

[@jeffersonluizbento20]

Brilliant

[@thetimetraveller7833]

Can you find x without the information of "red is equal to blue" ?

[@HedgePork]

No. Or rather there are infinitely many solutions, as shown at the beginning where Andy shrinks the red square and grows the blue triangle - those are all possible values of x. So you need some additional constraint which only one of those infinitely many solutions will also fulfill.

[@xaikirox]

first time ever I solved the question by myself :D

[@starwaItz]

make a book

[@Dazzle_Novak_]

(20-x)x+(20-x)(10-x)=2x^2 20x-(x^2)+200-10x-20x-x^2=0 -2(x^2)-10x+200=0 D=100+1600 x1=(10-sqrt(1700))/-4=7.808 x2=(10+sqrt(1700))/4=-12.808

[@obtusefool]

HOW exciting

[@Braden-York]

Gtp4 could not find x.

[@DarkReaperEdits]

i did the longer and worse way by adding up all the components to 200 (got the answer tho!)

[@alpha.4fx]

Thanks for the -ve value of x.

[@taltew]

Best birthday present ever, how exciting

[@rigs_halfcabs]

happy birthday taltew

[@samarthchauhan3866]

Happy Birthday !!!! :D

[@Nova1006]

Happy birthday :)

[@AndyMath]

Happy Birthday!

[@samarthchauhan3866]

@@AndyMath Hello Mr. Andy

[@eFse7en]

Is the other solution valid? Yes, but no 😎

[@rdspam]

Stream of thought… X2 = 5*(20-x) X2 = 100 - 5x. X2+5x-100=0 [-5 +/- sqrt(25 + 400)]/2 (-5 +/- 5sqrt17)/2. (5sqrt17-5)/2

[@MrMike3137]

Rock on

[@samarthchauhan3866]

Nice👍

[@Nova1006]

Nice

[@samarthchauhan3866]

@@Nova1006 Nice

[@lefthandedscout9923]

Huh, I really overcomplicated this one. My method was to say that the rectangle's area is 200, and we can work out the area of the right angle triangle to be 1/2 of bh so x/2(20-x) and the trapezium (I think that's the name of it? Idk) to be 1/2 of (a+b)h so 1/2(20+x)(10-x) then we know the red square's area is x^2 and the triangle's area as x^2 too, so them added together makes 200, or 2x^2 + 10x - x^2/2 + 1/2(200 - 20x + 10x - x^2) = 200, so 2x^2 + 10x - x^2/2 + 100 - 10x + 5x - x^2/2 = 200, which subtracted out gets x^2 + 5x + 100 = 200, take away 200 from each side for x^2 + 5x - 100 = 0, then solve with the quadratic formula. Still gets the right answer, just with 2 extra steps! I also immediately noticed the outside the box square the second I saw the question before even starting the solution. I've got an eye for silly answers that are technically correct in maths like that which my teachers never liked as it showed I thought of something they didnt, LOL