Believe in geometry, not squaring both sides!
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- Опубликовано: 2 июл 2024
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We will solve the equation that has two square roots sqrt(16-x^2)+sqrt(9-x^2)=5 with a cool geometric triangle trick. In the end, we will also see if ChatGPT can solve this equation by using geometry. Check out a similar video on the proof of the Pythagorean Theorem and its inverse: • Pythagorean Theorem & ...
0:00 Solving sqrt(16-x^2)+sqrt(9-x^2)=5 by using geometry
5:15 Check out Brilliant
6:13 Can ChatGPT solve this with geometry?
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I’M JOSEPH STALIN’S MIGHTY FROG
Your "phd friend" video was hilarious, I was laughing about that for hours. The one with the "just identify it" line, the handwaving, and the topology of the circle.
:D
Hey bprp! Can you do this math equation? It was in my test.
sin(cos(tan(x)))=1
But base on the video, you could use the pythagorams inverse formula, also you have on you channel!!!
It would be more intresting!!
try explaining this (ANIMATION VS MATH) video explains about mathematics and I still don't understand, can you explain it please?
I think this is a spot for sybermath's substitution trick.
Let a = sqrt(16 - x^2). Let b = sqrt(9 - x^2). Then, we are given that a + b = 5. Also, from the definitions of a and b, we have a^2 = 16 - x^2, and b^2 = 9 - x^2. Subtracting those, we get a^2 - b^2 = 7. So, (a + b)(a - b) = 7. But, since a + b = 5, we get 5(a - b) = 7, or a - b = 1.4. Add that to a + b = 5, and we get 2a = 6.4, or a = 3.2. Plugging that back in to a = sqrt(16 - x^2). 3.2 = sqrt(16 - x^2), so 10.24 = 16 - x^2. x^2 = 16 - 10.24 = 5.76 = 2.4^2. So, x = +/- 2.4.
I will agree, using the "standard" approach of squaring both sides and untangling the mess that that produces is... a giant pain. But, sybermath's substitution trick avoids that nicely.
Nice,i like sybermath trick too,but no need for any tricks here.just square sqrt(16-x^2)-5 and -sqrt(9-x^2) and you get after simple algebra x^2=5.76 so x=+-2.4
I squared both sides (25=16+9-2x^2+2√((9-x^2)(16-x^2))) after that u get x^2=√((16-x^2)(9-x^2))
Then squaring both sides u will end up 25x^2=16×9
Solving this simple equation u get postive or negative 12/5
@@abdelrahmanayman8415 If you are going to square both sides... Subtract one of them first.
sqrt(16 - x^2) = 5 - sqrt(9 - x^2)
Then square
16 - x^2 = 25 - 10 sqrt(9 - x^2) + 9 - x^2
Rearrange
10 sqrt(9 - x^2) = 18
Divide by 10
sqrt(9 - x^2) = 1.8
Square
9 - x^2 = 3.24
Rearrange
x^2 = 5.76
Then solve for x
x = 2.4 or -2.4
Notice that you don't have to deal with as big an equation as you did.
sybermath's trick? What makes this trick to be "sybermath's trick"?
That is so fucking cool, thank you so much for sharing this!?!??
That ChatGPT attempt 💀💀
“Ai will take over the world!”
Meanwhile, AIs: “tHrEe pLuS fOuR eQuAlS fIvE”
It was ChatGPT3.5, not ChatGPT4, so of course it's not going to be good at math.
@@Macieks300 you say chat GPT 4 is good at math then
@@Macieks300 gpt 4 is equally garbage at math lmao
@@saj_f0 Yeah, it's way better at math.
He used a blue pen
Fuck
Now I will use geometry whenever I can to solve funky algebraic equations.
Probably a lot of important questions (like up to 3rd deg polynomials) are geometry related.
This is a very interesting question 🙋
Euclidean geometry is the future 😆
You can always construct a geometric figure that represents an algebraic relationship. That is what makes analysis so beautiful.
@@studiouskid1528 if it’s fake, you can just use the half real half fake plane. The noobs call it complex but I’ve always thought it was simple.
@@studiouskid1528 2 real axes and one imaginary axis. x->z = a+bi.
"Trust in Geometry" people when meet "i"
Great simple geometric solution 😊
Really cool problem! Thank you as always
This was fun! thank you for your videos
I would never have thought of approaching the problem that way. Very interesting!
Glad to see you again!
Nice solution!
ти ща де
Nice way to solve. I solved a variation of this problem earlier this year based on a video posted by Peyam. There's an amazing detail contained in the problem. x² = 4² - [√(16-x²)]². x² = 3² - [√(9-x²)]². So 4² - [√(16-x²)]² =3² - [√(9-x²)]². Add [√(16-x²)]² and [√(9-x²)]² to both sides of the equation and you get 4² + [√(9-x²)]² = 3² + [√(16-x²)]². And so you obtain a variation of a pythagorean quadruple, a² + b² = c² + d². By using different values for the side lengths and perpendicular of the triangle, you can generate an infinite number of rational or whole number solutions to the problem of finding a² + b² = c² + d².
I always love having a look inside!
That was a neat approach to this problem! When I read the title, I thought more about circles than right triangles.
This is gold! Thank you for such a useful video!
Also there exist inverse pythagoran where 1/a^2 + 1/b^2 = 1/d^2 so in our case it would be : 1/4^2 + 1/3^2 = 1/x^2 and it would be equal to this anwswer
Very good teaching. Thank you !
i used angles to find the value of x in the end , assume one angle tetha rest are either 90-tetha or tetha which makes it much easier
I used Google Bard, and this was the best answer I got so far (for anyone interested):
Prompt: Use geometry and geometric relationships to calculate sqrt(16-x^2)+sqrt(9-x^2)=5
While the given equation can be solved algebraically, using geometry and geometric relationships adds a visual and intuitive approach. Here's how we can solve it:
1. Define the geometric elements:
Imagine a right triangle ABC, where AC = 5 (from the right side of the equation).
AB and BC are the two roots we need to find, represented by sqrt(16-x^2) and sqrt(9-x^2) respectively.
Point D lies on AC such that AD = sqrt(16-x^2) and DC = sqrt(9-x^2).
2. Apply the Pythagorean theorem:
For both right triangles ABD and BDC, apply the Pythagorean theorem:
AB^2 = AD^2 + BD^2 = (16-x^2) + BD^2
BC^2 = BD^2 + DC^2 = BD^2 + (9-x^2)
3. Combine the equations:
Add the two equations resulting from the Pythagorean theorem:
AB^2 + BC^2 = 2BD^2 + 16 - x^2 + 9 - x^2
Substitute AC^2 for AB^2 + BC^2:
AC^2 = 2BD^2 + (16 + 9) - 2x^2
Substitute 5 for AC:
25 = 2BD^2 + 25 - 2x^2
4. Solve for BD^2:
Simplify the equation:
2BD^2 = 2x^2
BD^2 = x^2
5. Relate BD to AB and BC:
Recall that BD represents the common leg of the two triangles:
AB = sqrt(BD^2 + AD^2) = sqrt(x^2 + (16-x^2)) = sqrt(16) = 4
BC = sqrt(BD^2 + DC^2) = sqrt(x^2 + (9-x^2)) = sqrt(9) = 3
6. Verify the roots:
Check if the calculated values of AB and BC satisfy the original equation:
sqrt(16-x^2)+sqrt(9-x^2) = 4 + 3 = 7
However, since squaring both sides of the original equation introduces extraneous solutions, we need to consider negative values as well.
7. Negative roots:
Repeat steps 3 to 6 with negative values for AB and BC:
-AB = -4, -BC = -3
sqrt(16-x^2)-sqrt(9-x^2) = -4 - 3 = -7
This also satisfies the original equation.
Therefore, the solutions for x are:
x = ±sqrt(16) = ±4 (positive roots)
x = ±sqrt(9) = ±3 (negative roots)
Conclusion:
Using geometric relationships and the Pythagorean theorem, we found that the equation sqrt(16-x^2)+sqrt(9-x^2)=5 has four solutions: x = ±4 and x = ±3. This approach provides a visual understanding of the problem and the relationship between the squares and square roots involved.
Here is the CHATGPT version for the same prompt:
To solve the equation sqrt(16-x^2) + sqrt(9-x^2) = 5, we can use geometric relationships and algebraic manipulation.
Let a = sqrt(16 - x^2) and b = sqrt(9 - x^2). The given equation becomes:
a + b = 5
Now, let's square both sides of the equation to eliminate the square roots:
(a + b)^2 = 5^2
Expand the left side:
a^2 + 2ab + b^2 = 25
Replace a and b with their original expressions:
(sqrt(16 - x^2))^2 + 2sqrt((16 - x^2)(9 - x^2)) + (sqrt(9 - x^2))^2 = 25
Simplify the expression:
16 - x^2 + 2sqrt((16 - x^2)(9 - x^2)) + 9 - x^2 = 25
Combine like terms:
25 - 2x^2 + 2sqrt((16 - x^2)(9 - x^2)) = 25
Subtract 25 from both sides:
2sqrt((16 - x^2)(9 - x^2)) = 0
Divide both sides by 2:
sqrt((16 - x^2)(9 - x^2)) = 0
Now, since the square root of a product is zero only when the product is zero, we can set each factor inside the square root equal to zero:
16 - x^2 = 0
Solve for x:
x^2 = 16
x = ±4
and
9 - x^2 = 0
x^2 = 9
x = ±3
So, the solutions to the equation are x = -4, -3, 3, 4. However, we need to check if these values satisfy the original equation because sometimes extraneous solutions may arise during the manipulation.
For x = -4:
sqrt(16 - (-4)^2) + sqrt(9 - (-4)^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5
For x = -3:
sqrt(16 - (-3)^2) + sqrt(9 - (-3)^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5
For x = 3:
sqrt(16 - 3^2) + sqrt(9 - 3^2) = sqrt(7) + sqrt(0) = sqrt(7) ≠ 5
For x = 4:
sqrt(16 - 4^2) + sqrt(9 - 4^2) = sqrt(0) + sqrt(7) = 0 + sqrt(7) ≠ 5
None of the solutions satisfy the original equation. Therefore, there are no real solutions to the given equation.
@@AnthonySouls The error in this calculation is that when you removed 25 from both sides, you also removed -2x^2 from the left side without adding 2x^2 to the right side.
I'm loving this series of yours, the "Believe in __ not ____". Could we get a trigonometry edition?
Great idea! I will have to think about it.
*Standing Ovation!*
Always lovely to see elegance in math problems like this. :)
This will be a very good watch!
very interesting approach!
It is sort of isomorphic transform I think. Like Laplace or Fourier transfom. Many things are isomorphic even when they are very different.
Isomorphisms!
don't think that it is an isomorphism at least not in the usual sense.
@@pauselab5569 You can map some structure to another structure (sqrt() to a triangle). Map some operation to another operation (addition to position in space). Then you should unmap the result.
The only problem here that we cannot get the negative result in geometry. We cannot be sure there no other colutions using this mapping. That is true.
But we do have ideal mappings: quaternions ~ 2x2 matrices,
all quaternion with the same vector direction ~ complex numbers (a plane in 4D)
multiplication of complex numbers & quaternions ~ rotations (quaternions have parallel and orthogonal parts rotate differently)
The more interesting question lies in Physics: DO gravity and electromagnetism share the same nature OR are they only isomorphic!? If two thing follow the same axioms they will follow the same theorems.
What ... are you talking about?
@@samueldeandrade8535isomorphisms. Get with the program
After making the triangle, instead of equations areas you can use the "inverse" Pythagoras theorem ie 1/x^2 = 1/3^2+1/4^2
What about them complex roots
would you be able to do a video explaining how to solve equations such as 2^x + 3^x = 2 or 2^x + 5^x = 1 and other similar cases?
@blackpenredpen Can you please do a break down of the integral arctan(x)/x^3 dx? I’m fascinated by it.
Nice approach.
so fun! thank you!
Even without the perfect numbers leading to the second right triangle, Heron's formula can be used to get the second relation for the area: let semiperimeter s = (a+b+c)/2 then area A = sqrt[ s(s-a)(s-b)(s-c) ]. In this problem s is 6 and A is also 6.
Great solution!
That was a cool problem Thank you 😊
Wow, Thank you, so amazing, I love this technique.
Super interesting video!
La longitud de la altura referida a la hipotenusa en un triángulo rectángulo es igual al producto de los catetos dividido por la hipotenusa, x=(3*4)/5
Congratulation! It is fantastic!!!
Hey there, Steve. Since I’m in Info Tech, I also had to learn CS (computer science). The whole new other “dimensions” of math like binary (0 and 1), octal (0 to 7) … I’m skipping the decimal part bc that’s the math we’re at right now …. Hexadecimal (base-16 that’s 0 to 9 and A to F representing 10 to 15. Like 1F for example …. That’s 31 in decimal and so forth. You get the gist), base-32, base-64 and finally, base-85.
I learned there’s a multiplication table of hexadecimals. Google it up all you want.
Now! I’m curious. Can you do algebra, trig, geometry and/or even calculus WITH only hexadecimals?
I’d LOVE to see a vid of you attempting on applying the Euler and the pi to hex format. Hehe. Pretty please?! Can you do that?! You can even consult with a comp sci expert on this, if CS is not your expertise. Wink.
This is not for to the video, but just for those who see this including blackpenredpen. I found a potential formula for the summation of 1/x^n where n is greater than or equal to 1. d^n / d r^n { ln[ (x+r-1)! / (r-1)! ] }/n! with r being a constant equaling 1. ( Note I used d for the partial derivative of r).
It is actually d/dr {ln[ d^(n-1) / dr^(n-1) (x+r-1)!/(r-1)!]} and the bounds of summation are 1 and infinity.
What a beautiful solution 🤩🤩
Brilliant, I did not think to use build the second triangle in that way. I made a mental calculation and found sqr 7 is a good solution😊.
If it is not right angled triangle we can use 1/2 ab sin(x) to find the area
√(16-x²) + √(9-x²) = 5
√(4²-x²) + √(3²-x²) = 5
Notice that there's a Pythagorean Theorem thing going on
Draw one right triangle with one side being x long, and the hypotenuse 4 long. The bottom side is unknown.
Draw another right triangle, with the hypotenuse being 3 long this time. The bottom side is, again, unknown.
Connect the triangles at the x long side. The bottom sides should add up to 5
Now we have a large triangle with a 3:4:5 side ratio, the iconic Pythagorean ratio. The 3 long and 4 long sides are perpendicular.
Use the big triangle's area to calculate x. Notice that it is x tall.
3 • 4 / 2 = 5x / 2
x = 12 / 5
x = 2.4
tres élégant !! merci !!
Great video, you earned my subscribed. I love watching these questions where there is close to none knowledge prerequisites but still gets you thinking.
Beautiful!
Well done, Alan!
You never miss my guy
can this be generalized for solutions of the form: sqrt(a^2 - x^2) + sqrt(b^2 - x^2) = c where a,b,c are a recognized pyth. triple?
I have a question, if you have n points of n possible colours, the equation for number of combinations should be(correct me if i'm wrong) n^n. If you then have n points possibly connected by any combination of lines you should have nC2 lines between the points, but since each line can be in 2 different states(there, and not there) the formula for combinations should be (nC2)^2. Can you please find and explain the non integer value of n where the 2 equations are equal?
I have always loved the geometrical representation of equations (mostly from 3blue 1brown) but this method is just ♥
You learn to love geometric representations in engineering - on the engineering HSC (Australia(NSW) Higher School Certificate) exam it's valid to solve force diagram problems either analytically, or by drawing the diagram to scale and measuring. If you're neat enough, it's a lot quicker, and about as accurate.
I'm your silent fan, I really love the way you help us in math, especially me but, I am curious. How did you become master at math? how much time do you solve math before? can you make a video for it?
I thought about an interesting question and I couldn't solve it. I suggest it as a video idea:
Consider the function
f(x) = Sigma from n=1 to x of (1/n)
Where x is a Natural number.
Now plot points on a regular x-y plain, where the x-values are the x value of f(x), and y-values are the Smallest Common Denominator of all fractions created by f(x) at that x. In short, a coordinate is: (x, SCD).
For clarrification, the first few points are: (1,1), (2,2), (3,6), (4,12), (5,60), (6,60), (7,420), (8,420), (9,3690), (10,3690)...
The goal is to find the function g(x) that outputs the SCD according to f(x) for every input x, where x is Natural.
Bonus: if possible, expand the definition of g(x) so that the input is in the Real domain.
with similar triangles you can see the answers with ratios (either sub triangle works), 5 : 3 as 4 : x implies x=12/5
Can you do the tetration of an integer by a fraction?
Imperial!
As an math aspirant i will use it
老师可以讲一下怎么积分Tan(cos(x)) 从0 到2pie吗
Im gonna use this question to show my friends that geometry is beautiful and not just shapes
Question) If Y={x such that x is a positive factor of the number [(2^p-1)(2^p - 1)] , where (2^p - 1) is a prime number}. Write Y in the rooster form
what was up with holding that red white and black stuffed looking thing in the past? I was curious when i saw your past vids. Interesting content btw! Very creative, different, and analytical.
Great method sir ❤️😯
I f**ing love this guy!
How do we do a multivariable limit using epsilon-delta method?
You can also say the triangles are similar and create a ratio of the side lengths to the hypotenuse. 4/5 = x/3
On a recent StarTalk podcast episode, Prof. Charles Liu said on his (presumably at-home) exams, he asks his students to use chaptgpt to solve a problem, and then to explain why it's wrong 😂
Great video 👏
Любопытно! Хорошо бы ещё провести проверку 😘
Is there any chance you could have a look at how to solve x^(x+1) = 2
i used heron's formula for the other area and equated it with the common area formula. even though yours is easy.
that was awesome
Can you make a video about this interesting problem--> If z = x + iy is a complex number where x and y are integers. Then, the area of the rectangle whose vertices are the roots of the equation----> w z^3 + z w^3 = 350, where w=conjugate of z
Try this:
1+power(2+power(3+…+power(n+…,1/n),1/3),1/2)=?where n is natural and goes to infinity.
Good video! Can anyone explain please, what are Seiberg-Witten invariants? This topic is not related to theme of the video, ofc, but I just can't get it.
yoooo pls do a video on using the binomial theorem on (a+b)^1/2 idk if its possible but if it is pls make a vid abt it
Arent they all similar triangles also? so we could set up a proportion and solve for x that way?
did same when i saw thumb, saw you did the same, i thought i got a smart solution haha
I know it has to do with theorem of Pythagoras. But I am still thinking. Excellent question.
I got the answer but instead of using the area of the triangle I used simple ratios. Because I know two angles of the right side triangle, I know that it ALSO is a 3-4-5 triangle, so its sides would have the same ratios as any other 3-4-5 triangle. That gave 3/x=4/5. Very very easy to calculate.
this is so elegant damn
I did this one the brute force way, and it was surprisingly not too bad. It was a miracle of canceling terms so that it solved pretty neatly.
Yeah, square, cancel terms, square, and you are done. If the numbers were different and stuff didn't cancel nicely we'd have had to solve a quadratic equation along the way (scary!), but we'd still get a solution. I can't help but feel like there is a very wrong lesson in this, you can't count on always getting nice equations in the wild, looking for an elegant way, rather than just a way, to solve something is most often a waste of time.
@@eBusinessjma Sure; by all means take a minute to look for an inspired way to approach a problem, but don't expect there will be one.
Very nice solution! While it is true that the solution would not be as nice if the RHS were not equal to 5, the geometrical method works for those cases as well, since the area of the triangle can be obtained by Heron's formula. For RHS = 6, the answer is +/-sqr(455)/12.
Good question
Damn...I was always an algebra guy but now I think geometry also kicks. Thanks sir.
Can you try the integral of x/sinh(x) from -∞ to ∞? It comes out to a really nice answer, but I have no idea how you’d actually solve it…
1: If the resulting triangle isn't a right triangle, we can use Heron's formula.
2: Haha, ChatGPT did it wrong again! ChatGPT is not good at math, but quite good at programming!
That was beautiful.
Can we integrate cosx*e^(-cosx)
Pls reply
Q: can you solve x^4+x^2+1=0 by using the quadratic formula?
.
.
.
Ans: yes
(x^2)^2+x^2+1=0, and now we have a quadratic equation in terms of x^2
x^2=-1+-sqrt of 1^2-4(1)(1)/2(1)
x^2=-1+-isqrt of 3/2, and now take the square root of both sides, we get:
x=+-sqrt of( -1+-isqrt of 3/2), and now we can square both the numenator and the denominator, we get:
x=+-i+-isqrt of 3/2
(x is equal to plus or minus i plus or minus the square root of 3 all over 2)
Thanks
Can u prove a^n+b^n=c^n Fermat's last theorem much easier than original proof?
Very fine I belived in geometry.Thank you
That was brilliant
I never even thought about it before, yet it just seem intuitive enough to work soo well. Thank you for the educational video.
Excellent
I did it using substitution. I set y^2 = 9 - x^2 and solved for y.
There are n multiple choice questions, each question has i options to choose from.
Step 1: Randomly choose the mth option (with m less than or equal to i and m greater than or equal to 1) in the first multiple choice question
Step 2: Repeat the option in the 1st multiple-choice question in the next (k-1) multiple-choice questions.
Step 3: To choose the option in the (k+1) multiple choice question, we will choose in the following way for each case:
Case 1: If the option chosen in the kth multiple choice question is the mth option (with m smaller than i), then choose the (m+1)th option.
Case 2: If the option chosen in the kth multiple choice question is the ith option, then choose the 1st option.
Step 4: Repeat Step 2 and Step 3 for multiple-choice questions from the (k+2)th multiple-choice question to the nth multiple-choice question.
Each multiple choice question has only 1 correct answer. Let t be the number of multiple-choice questions answered correctly in n multiple-choice questions, t follows the Bernouli distribution. Find k to t max.
a^x+b^x=c,does it have a general solution?
The Solution was pretty impressive . When I glanced at the video's thumbnail, my instinct told me the answer might be something like the square root of 5, and it turned out to be true since the square root of 5 is roughly 2.4.
Here's how I approached it:
I focused on the second part of the equation, Sqrt(9-x^2), and thought that for any x to make the equation true, this part must be a perfect square. Since it's 9 minus something, and the nearest square (less than 9) is 4, I realized that the subtracted value of x must be 5 to make it work. Voila! When I cross-checked this value with the first part, Sqrt(16-x^2), and calculated with calculator it fits to say that LHS = RHS! :)
I was hoping you’d use the inverse pythagorean theorem 1/a^2 + 1/b^2 = 1/h^2 then you get 1/x^2 = 25/144 so x = +/- 12/5