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 ..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
Negative solution interpretation blew my mind, the placement changing based on negative/positive options and the areas still being equal was very exciting indeed.
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.
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.
@@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!
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!
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!
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
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.
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.
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.
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
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.
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.
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.
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.
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😂
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.
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
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.
@@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.
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)
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.
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.
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?
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
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
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.
what. this is literally unsolvable, like he shows the triangle(along with the square) can have ANY size. It would not affect the big rectangle in any way. There is no solution to this, this is literal theatrics with some far off assumptions. X could be ANY size between 0.(0)1 and 9.(9) and there is no way to actually measure it. what is this even about. What do you mean solution. This is UN-solvable.
