Lot of confusion in the comments about the p and q part. Quick written explanation for anyone not sure what happened in that step, it can be reduced from a quartic to a simple quadratic with a substitution of a=x². Solving a² - 78a + 1296 = 0 using the quadratic formula gives the roots.
It might be worth noting that before we dive into the algebra, we can infer some properties of the solutions. We are trying to solve two simultaneous equations whose solutions correspond to the intersection of a circle of radius SQRT(78) and a hyperbola with vertices at (6,6) and (-6,-6); we know that any solutions will be mirrored in the first and third quadrants of the cartesian plane. We can also infer that since x and y are interchangeable in the equations, they must be so in the solutions (i.e., you should be able to swap the x & y values of a solution to get another solution - another way of saying the solutions are reflected in the line y = x). Finally, we can infer there must be 4 real solutions (two in each of quadrant 1 and quadrant 3), rather than zero or 2 (one in each quadrant), by noting that the circle’s radius is greater than SQRT(2*36), so it must intersect the hyperbola at two points in both quadrants (draw yourself a diagram of the two equations in the cartesian plane to see this). If the radius was less than SQRT(72), there would be no intersections (i.e., zero real solutions), and if it was exactly SQRT(72) (i.e., x^2 + y^2 = 72), then there would be exactly one solution in quadrant 1 and one in quadrant 3 (i.e., at the hyperbola’s vertices). Of course, knowing there are two solutions in each quadrant doesn’t help you figure out their values, but at least you can be sure they are there to be found.
Maybe a substitution in the quartic equation where you had x^4 and x^2. Set something like u=x^2 and solve with the quadratic formula for u. And then you can do it again for x?
Hi &E, love your math puzzles! Just wanted to point out that the answers could be simplified and the y values look different but are equal to the alternate x values. 36/√54 = √24 and 36/√24 = √54. Doing the math to prove that they are the same was kinda fun. But the possible solutions in simplest form are: X=2√6 Y=3√6 or X=3√6 Y=2√6 or X=-2√6 Y=-3√6 or X=-3√6 Y=-2√6 Not trying to be a know it all because I'm not, if your puzzle involves circles my solve rate is about 50% without hints, my trig is very rusty! I look forward to each new math puzzle, keep up the great content!
I'm convinced that most of us watch Andy because we love math. But, us women (some men, too) watch Andy also because he's cute. Sorry that my comment is so long. (My husband and I quote Gene from Bob's Burgers all the time, "we're married, not buried." There is no shame in admiring beautiful people; that's why while rewatching Home Improvement I had to skip the episode where Jill gets super mad and jealous because Tim checked out another woman). 😆
@Andy Math I don’t know if it will be easier to use identity for this question since for this way it doesn’t nneed to find the solution of x and y and we can calculate to the final answer ?
Sorry for bad English but it's a known condition for cases such as that one which let you simplify things but not everyone in the world learns about them. Like how some learn about the ABC formula but not PQ formula and vice versa.
(x+p)(x+q)=x²+px+qx+pq, pq is a constant with no x, therefore the two numbers we are trying to find, when multiplied, should equal to the constant. In this case, 1296. px+qx means that the coefficient of x is p+q, therefore p+q should be equal to the coefficient of x. E.g. factorisation of x²+2x-3, p+q = coefficient of x, which is 2 pq = constant, which is -3 You then do trial and error till you find the correct combination. In this case, the answer is (x+3)(x-1)
@@AkitoLitethanks, i really wasn't in the mood to think about it, but 2 people came in to help, so i forced myself to rewatch the video and think about it with your explanation i do feel like i understand it better, specially due to that example with x² +2x -3, the last part having (x+3)(x-1) made me feel like i maybe understand it now, thanks
So are you telling me that we’ve seen amazing videos like this for more than 7 years and we’ve never had a video to know more about Andy? Why Andy? Why?
As to solving this problem using a longer method, Good Job - easily understood and not really difficult. As to the Haircut, Good Work there, too! Although, I don't understand why someone with great hair needs to have a hat on indoors and in front of a camera? 😂😂😂
hey Andy great video, however I had an alternate solution to this, just square on both sides the equation x²+y²=78 and we'll have the answer without having to deal with square roots 😅
I would like to understand the p and w part more, I feel like when that started happening in math class it really took a lot of joy out of math for me because my brain does not comprehend
Sometimes factoring is too difficult or takes more time than to use the quadratic formula. The quadratic formula works here because if you subsitute x^2 for let's say u, then it becomes a quadratic equation. Then you can use the quadratic formula, and then you can substitute the x^2 back into u. Hope this is clear and helpful!
Whenever you have a formula that us written like: a^2 + 2ab + b^2 You can transform it into a format (a+b)(a+b) This is easy because b can always be found with root of b^2 The tricky part is knowing what the factors are when the formula is: a^2 + a(b + c) + bc Which is (a+b)(a+c) How is thus second method done you might ask? Well, lets make an example: x^2 + 10x + 16 1. First you look at which numbers factor up to x^2 - that's simple it's x - so we fill it in the formula: (x+b)(x+c) 2. You look at which numbers factor up to 16 - 4*4 = 16 - 16*1 = 16 - 2*8 = 16 - etcetera 3. Now you look at which of these would summ up to 10: - 4+4=8 - 16+1=17 - 8+2=10 4. so the formula is (x+8)(x+2) Now things can get more complex when using minus signs or fractions, but let's not get ahead of ourselves. You can test your skills online researching: trinomials or binomial products.
Hi! I dont really know where i am! I was wondering youtube and went into my youtube channel, and saw that you are a subscriber of mine :) How did that happen?
I want to add a solution for my side Add and subtract the eqn 1 by 2 x^2y^2 So the eqn become x^4+y^4 +2x^2y^2 -2x^2y^2 Then by identity a^2+b^2 +2ab= (a+b)^2 (X^2+y^2)^2 -2x^2y^2 Put the value (78)^2 - 2 (36)^2 6084- 2 (1296) 6084 -2592 3492
Fun fact: Andy is always handsome with any hairstyle
Yep. When he says "how exciting" I always think he is talking about himself.
I see why Andy left Toy Story
gay
@@burntsouffleoms 😭🙏
@@leodame3 omsimize
Let's get an hour long live stream of Andy solving fan submitted math problems!!!
YES PLS3AS3
Not just exciting, absolutely electrifying.
How exciting ❌
How absolutely electrifying ✅
Target audience 🧒
Actual audience 🧑💼
For real! I need none of this math. I'm an accountant. 😆
Lot of confusion in the comments about the p and q part.
Quick written explanation for anyone not sure what happened in that step, it can be reduced from a quartic to a simple quadratic with a substitution of a=x². Solving a² - 78a + 1296 = 0 using the quadratic formula gives the roots.
It might be worth noting that before we dive into the algebra, we can infer some properties of the solutions. We are trying to solve two simultaneous equations whose solutions correspond to the intersection of a circle of radius SQRT(78) and a hyperbola with vertices at (6,6) and (-6,-6); we know that any solutions will be mirrored in the first and third quadrants of the cartesian plane. We can also infer that since x and y are interchangeable in the equations, they must be so in the solutions (i.e., you should be able to swap the x & y values of a solution to get another solution - another way of saying the solutions are reflected in the line y = x). Finally, we can infer there must be 4 real solutions (two in each of quadrant 1 and quadrant 3), rather than zero or 2 (one in each quadrant), by noting that the circle’s radius is greater than SQRT(2*36), so it must intersect the hyperbola at two points in both quadrants (draw yourself a diagram of the two equations in the cartesian plane to see this). If the radius was less than SQRT(72), there would be no intersections (i.e., zero real solutions), and if it was exactly SQRT(72) (i.e., x^2 + y^2 = 72), then there would be exactly one solution in quadrant 1 and one in quadrant 3 (i.e., at the hyperbola’s vertices). Of course, knowing there are two solutions in each quadrant doesn’t help you figure out their values, but at least you can be sure they are there to be found.
That's a great illustration thanks!
Short cut, long path. Whats the medium solution? Lol
Maybe a substitution in the quartic equation where you had x^4 and x^2. Set something like u=x^2 and solve with the quadratic formula for u. And then you can do it again for x?
x2 + y2 - 2xy= 78 - 2*36
(x - y)2 = 6
x - y = sqrt(6)
x = y + sqrt(6)
plug it into xy = 36, find x and y
Is Andy the oldest 20yo or the youngest 40yo?
Maybe he is just 30......?
Astounding reasoning again by Mr Andy 👍
I do NOT get any of these videos but I still watch them 🥴
Same
fire haircut🔥
Neat solution and neat haircut.
Takes hat off.
I’ve got hat hair now, I don’t know why.
😂
which program you use to make that motions with the equations? I would love to use it in my classes
Hi &E, love your math puzzles! Just wanted to point out that the answers could be simplified and the y values look different but are equal to the alternate x values.
36/√54 = √24 and
36/√24 = √54.
Doing the math to prove that they are the same was kinda fun.
But the possible solutions in simplest form are:
X=2√6 Y=3√6 or
X=3√6 Y=2√6 or
X=-2√6 Y=-3√6 or
X=-3√6 Y=-2√6
Not trying to be a know it all because I'm not, if your puzzle involves circles my solve rate is about 50% without hints, my trig is very rusty! I look forward to each new math puzzle, keep up the great content!
math student: QED
andy math student: how exciting 🤩
Andy math is the type of guy that gives my 10th grade math knowledge a purpose 😂
I'm convinced that most of us watch Andy because we love math. But, us women (some men, too) watch Andy also because he's cute.
Sorry that my comment is so long.
(My husband and I quote Gene from Bob's Burgers all the time, "we're married, not buried." There is no shame in admiring beautiful people; that's why while rewatching Home Improvement I had to skip the episode where Jill gets super mad and jealous because Tim checked out another woman). 😆
Thank you, Andy.
Cool I found the same result but with a different way❤❤🎉. How exciting 🎉🎉🎉🎉
Love this guy!
@Andy Math I don’t know if it will be easier to use identity for this question since for this way it doesn’t nneed to find the solution of x and y and we can calculate to the final answer ?
HOW EXCITING 🔥🔥🔥
Always exciting solutions!
Hello Andy, I'm a fan of your content and I'd like to suggest a math problem:
fully simplify: 3^100 + 3^100 + 3^100/3^101-3^100-3^99
Well As far as I could go,
2×3^99 + 1/3
i didn't really understand that part with p and q, i probably could if i think about it long enough tho
Sorry for bad English but it's a known condition for cases such as that one which let you simplify things but not everyone in the world learns about them. Like how some learn about the ABC formula but not PQ formula and vice versa.
(x+p)(x+q)=x²+px+qx+pq,
pq is a constant with no x, therefore the two numbers we are trying to find, when multiplied, should equal to the constant. In this case, 1296.
px+qx means that the coefficient of x is p+q, therefore p+q should be equal to the coefficient of x.
E.g. factorisation of x²+2x-3,
p+q = coefficient of x, which is 2
pq = constant, which is -3
You then do trial and error till you find the correct combination.
In this case, the answer is (x+3)(x-1)
@@AkitoLitethanks, i really wasn't in the mood to think about it, but 2 people came in to help, so i forced myself to rewatch the video and think about it
with your explanation i do feel like i understand it better, specially due to that example with x² +2x -3, the last part having (x+3)(x-1) made me feel like i maybe understand it now, thanks
pov: you have not developed dynamic problem solving skills
Loveee the hair 😂
Greetings from Poland.
greetings from kenya
I'm from Brazil and i really liked your videos. What programs do you use to make these videos?
Mostly PowerPoint
How exhilarating!
Fans: no shortcut
Andy: I did use the calculator.
“how exciting” indeed
Thats a pretty nice haircut
Let a = x^2,b = y ^ 2.
Therefore,from what's given,
a + b = 78 __ (1)
And we knew that xy = 36
Therefore,x^2.y^2 = 36.36 = 1296.
Hence,
ab = 1296.
Now x^4 + y^4 = a^2 + b^2,
Also, a^2 + b^2 = (a+b)^2 - 2ab
Also,
a+b = 78,ab = 1296
Therefore,(a^2 + b ^ 2) = (78)^2 - 2 * 1296
= 6084 - 2592
= 3492.
Worries about *multiplying* by x if it's zero (??), but has already blithely divided by x 🤔. The haircut makes up for everything though :-)
Could you not have solved the quadratic equation (where x square equals z for example) instead of looking for p an q manually?
The haircut was our present for watching the whole video
Maths are beautiful
How exciting🌟🤩
So are you telling me that we’ve seen amazing videos like this for more than 7 years and we’ve never had a video to know more about Andy? Why Andy? Why?
As to solving this problem using a longer method, Good Job - easily understood and not really difficult. As to the Haircut, Good Work there, too! Although, I don't understand why someone with great hair needs to have a hat on indoors and in front of a camera? 😂😂😂
hey Andy great video, however I had an alternate solution to this, just square on both sides the equation x²+y²=78 and we'll have the answer without having to deal with square roots 😅
HOW EXCITINGGGGF
Also, if x=+-sqrt(24) THEN y=+-sqrt(54) and vice versa ;).
How exciting !!!
I think it can be solved using the Newton-Girard formulas
How exciting.
Something unrelated to Math. Can you make a video saying “how exciting” on a loop? Haha
X^2 on top and bottom reduce to 1! Not cancel. They are not positive and negative charges.
W haircut
I noticed (x^2 + y^2)^2 = (x^4 + y^4) + 2(xy)^2. From there its simple substitution and algebra
How exciting
😂 how exciting🎉
I always solved w the shortcut and thus never thought or tried the long* cut
Hello Andymath
czekolady w 3 min
I would like to understand the p and w part more, I feel like when that started happening in math class it really took a lot of joy out of math for me because my brain does not comprehend
I think there is a formula to solve it, but idk if it works
p+q = number 1
p=number1-q
(number1•q)(q) = number2
number1q q² = number 2
then solve it
Sometimes factoring is too difficult or takes more time than to use the quadratic formula. The quadratic formula works here because if you subsitute x^2 for let's say u, then it becomes a quadratic equation. Then you can use the quadratic formula, and then you can substitute the x^2 back into u. Hope this is clear and helpful!
Whenever you have a formula that us written like:
a^2 + 2ab + b^2
You can transform it into a format
(a+b)(a+b)
This is easy because b can always be found with root of b^2
The tricky part is knowing what the factors are when the formula is:
a^2 + a(b + c) + bc
Which is
(a+b)(a+c)
How is thus second method done you might ask? Well, lets make an example: x^2 + 10x + 16
1. First you look at which numbers factor up to x^2
- that's simple it's x
- so we fill it in the formula:
(x+b)(x+c)
2. You look at which numbers factor up to 16
- 4*4 = 16
- 16*1 = 16
- 2*8 = 16
- etcetera
3. Now you look at which of these would summ up to 10:
- 4+4=8
- 16+1=17
- 8+2=10
4. so the formula is (x+8)(x+2)
Now things can get more complex when using minus signs or fractions, but let's not get ahead of ourselves. You can test your skills online researching: trinomials or binomial products.
Hey Andy, I don't know if this is the place but do you have a degree in math?
How do I submit questions
Andy math op😊😊
how exciting
Hi! I dont really know where i am! I was wondering youtube and went into my youtube channel, and saw that you are a subscriber of mine :) How did that happen?
Andy how old are you
Guys, how can i send him a math problem?
To be honest, I wanted to see you show how you got p & q even though you used a calculator 😅
Andy, give my comment a heart, pls
I just use x² =54/24 to get y²=24/54
And then just put this in equation to het 3492 why would you go to sqrt x²?
Leaving your y with a sqrt in the denominator hides the symmetry between x and y :(
You are right. That would have been better to show that.
2592?
no views, one comment and three likes.. youtube is drunk again.
true..
false...
Some people like and comment before they have watched enough of the video where UA-cam will count it as a view.
Bro this is 8th grade problem damn.
Can anyone sharethe link for the shortcut method for these type of problem
Bro pls react and solve the que of iit jee advance maths
2 and 3 for x and y and 97 for the answer? At time 0:00 , used the thumbnail to answer, no calculator
You should tell to your barber to cut your hair without the shortcut.
Much much easier solution...
x² + y² = 78
x⁴ + 2x²y² + y⁴ = 78²
2x²y² = 2x36²
x⁴ + y⁴ = 78² - 2x36²
how is that a shortcut?
żb stratwgia lifestylu byla faktycz dobra
78^2 - 2(36^2)=3492
-24 and 24 sum to zero!
Normal Comment
ua-cam.com/video/k8pIAJeOPY4/v-deo.htmlsi=kzngn2K3YNzbUGtw
Is this even legal maths?
Have you ever heard of the formula a+b whole square 😒.. stupid method
Nerd
Too long. Short solution: 78^2 - 2* 36^2
It was too easy for me!
(x² + y²)² - 2(xy)² = x⁴ + y⁴
Put the values of x² + y² and xy
You will get your answer quickly
very long, shortcut method (a+b)^2 = a^2 + b^2 + 2ab here a,b are x^2 and y^2 respectively then (78)^2 = x^4 + y^4 + 2(36)^2
I want to add a solution for my side
Add and subtract the eqn 1 by 2 x^2y^2
So the eqn become
x^4+y^4 +2x^2y^2 -2x^2y^2
Then by identity
a^2+b^2 +2ab= (a+b)^2
(X^2+y^2)^2 -2x^2y^2
Put the value
(78)^2 - 2 (36)^2
6084- 2 (1296)
6084 -2592
3492