Can you solve for X? | (The Golden Ratio) |
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- เผยแพร่เมื่อ 8 ก.ย. 2024
- Learn how to find the value of X. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; Quadratic formula. Step-by-step tutorial by PreMath.com
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Can you solve for X? | (The Golden Ratio) | #math #maths | #geometry
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Another clear and succinct demonstration Professor !
Thanks for sharing Man !
Much appreciated!
You are very welcome!
Thanks for the feedback ❤️
Posting of your every sum is being enthusiastic as it contains minimum 3 concepts involving into it ....great sir
(x)^2+(x^2)^2=(x^3)^2
x^2+x^4-x^6=0
x^2(1+x^2-x^4)=0
x^2=0; x=0 rejected
Aand x^4-x^2-1=0
Let t=x^2 (t>0)
So t^2-t-1=0
So t^2=(1+√5)/2
x^2=(1+√5)/2
So x=(√1+√5)/2)..❤❤❤
Excellent!
Thanks for sharing ❤️
So the diagonal of a golden rectangle is φ^(3/2) times its width. I learned something new.
Nice introduction of golden ratio 🎉
Thanks for liking😀
Glad to hear that! ❤️
Beautiful step by step explanation. I prefer to teach the golden ratio using ratios. Also please put in that x=0 is a trivial solution so we assume x isn't zero and can divide. Nice use of substitution and radicals. In the USA the students need to be constantly reminded about the properties of zero. Very pretty problem.
Есть ещё интересное наблюдение - Х * Х(2)=Х(3). То есть диагональ прямоугольника равна его площади
Спасибо за отзыв ❤️
You can reject negative solution a bit earlier, when you split positive and negative solutions of quadratic equation. Why? Because x^2 is one of the sides of the triangle it HAS TO BE POSITIVE.
A great solution.
If it wasn't a triangle and was purely number crunching, I would have said solutions of 0 and 1. However, I also did some calculating and came up with x = the cubed root of 2. However, a further look seemed to indicate that x can be the cubed root of anything. I find this perplexing. I will look again later when I have more time.
Excellent!
Thanks for the feedback ❤️
Thanks for the unexpected challenge.
=> The triangle has for x= 1,25 cm, because it is the only realizable value, by using the faithful construction of the triangle as specified. The value of φ is not applicable here.
Thanks PreMath
Thanks Sir
That’s very nice
Enjoyable and useful method
With my respects
❤❤❤❤❤❤
thanks
Interesting question, here my solution ▶
for this right triangle ΔABC
[AB]= x²
[BC]= x³
[CA]= x
according to the Pythagorean theorem:
[CA]²+[AB]²= [BC]²
x²+(x²)²= (x³)²
x² + x⁴ = x⁶
x⁶ - x⁴ - x² = 0
x²(x⁴-x²-1)=0
x₁= 0 ❗(we know that x ≠ 0)
x⁴-x²-1 = 0
x²=u
⇒
u²-u-1=0
Δ= 1-4*(-1)
Δ= 5
u₁= (1+√5)/2
⇒
x²= (1+√5)/2
x₁= √(1+√5)/2
x₁= √1,618033989
x₁≈ 1,272 length units
u₂= (1-√5)/2
u₂= - 0,618
x²= - 0,618 ❗( x would be a complex number)
⇒
x= √(1+√5)/2
x ≈ 1,272
Excellent!
Thanks for sharing ❤️
x=√2•√(1+√5)/2≈1,272
Excellent!
Thanks for sharing ❤️
Ahh I got that but was hoping for an explanation of how to simply the square root of the golden ratio.
Good job!
Thank you! Cheers!❤️
Very easy.
Thanks for the feedback ❤️
X^3 = 1
X = +1 one of the answers....not aware of other 2
Let's find x:
.
..
...
....
.....
We have a right triangle, so let's apply the Pythagorean theorem:
(x³)² = (x²)² + (x)²
x⁶ = x⁴ + x²
x⁶ − x⁴ − x² = 0
x²*(x⁴ − x² − 1) = 0
Since x>0, we obtain:
x⁴ − x² − 1 = 0
x² = 1/2 ± √[(1/2)² + 1] = 1/2 ± √(1/4 + 1) = 1/2 ± √(5/4) = 1/2 ± √5/2 = (1 ± √5)/2
Since x²>0, we have only one useful solution:
x = √[(1 + √5)/2] ≈ 1.272
Best regards from Germany
Excellent!
Thanks for sharing ❤️
please tell some for Bangladesh🇧🇩.We are in trouble.I always see your video.
Our sincere thoughts and prayers are for you🙏
There is a hidden blessing behind every event. All the best dear❤️
What if a=x^3, b=x^2, c=x ?
Interesting question. Let's have a try:
(x)² = (x²)² + (x³)²
x² = x⁴ + x⁶
0 = x⁴ + x⁶ − x²
0 = x²*(x² + x⁴ − 1)
x>0:
x⁴ + x² − 1 = 0
x² = −1/2 ± √[(−1/2)² + 1] = −1/2 ± √[(1/4 + 1) = −1/2 ± √(5/4) = −1/2 ± √5/2 = (−1 ± √5)/2
x>0:
x = √[(√5 − 1)/2] ≈ 0.786
If c=x³ we obtain x = √[(√5 + 1)/2] = √φ.
If c=x we obtain x = √[(√5 − 1)/2] = √(1/φ).
Best regards from Germany
According to the Pythagorean theorem, we would have:
a²+b²= c²
(x³)²+(x²)²= x²
x⁶ + x⁴ = x²
x⁶ + x⁴ - x² = 0
x²(x⁴+x²-1)= 0
x₁ =0 ( x ≠ 0) ❗
x⁴+x²-1 =0
x²= u
⇒
u²+u-1=0
Δ= 1-4*1*(-1)
Δ= 5
u₁ = (-1+√5)/2
⇒
x²= (-1+√5)/2
x= √(√5-1)/2
x ≈ 0,78615
u₂= (-1-√5)/2
x²= (-1-√5)/2
x would be a complex number ❗
⇒
the only solution is x= 0,78615
Good question👍
I knew it!
It's the Golden Ratio
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
The Square Root of the Golden Ratio!!
@@LuisdeBritoCamachoYou got a sharp eye there. Nice!
Good night sir
Hello dear❤️
Likewise😀
x² + (x²)² = (x³)²
x² + x⁴ = x⁶
x⁶ - x⁴ - x² = 0
x²(x⁴-x²-1) = 0 --- x = 0 ❌
x⁴ - x² - 1 = 0
(x²)² - x² - 1 = 0
u² - u - 1 = 0 0
x = √((1+√5)/2) = √φ --- x > 0
x = √(1+√5)/√2 = √(2+2√5)/2 ≈ 1.272 units
Excellent!
Thanks for sharing ❤️
Does anyone know if the square root of phi ((1+sqrt5)/2) can be denested? I think probably not.
STEP-BY-STEP RESOLUTION PROPOSAL USING Phi - GOLDEN RATIO:
01) X^6 - X^4 - X^2 = 0
02) Change of Variable : Y = X^2
03) Y^3 - Y^2 - Y = 0
04) Y * (Y^2 - Y - 1) = 0
05) Y = 0 or Y^2 - Y - 1 = 0
06) Y = (1 - sqrt(5)) / 2 ; Y ~ - 0,618 (Negative Solution)
07) Y = (1 + sqrt(5)) / 2 ; Y ~ + 1,618 (Positive Solution)
08) As : Y = X^2
09) X^2 = 1,618 ; X = + sqrt(1,618) or X = - sqrt(1,618)
10) The Solution must be Positive!!
11) X = sqrt(1,618) ; X ~ 1,2720
12) Checking Solution : 1,2720^6 = 1,2720^4 + 1,2720^2 ; 4,236 = 2,618 + 1,6180 ; 4,236 = 4,236
Therefore,
OUR ANSWER :
X = sqrt(Phi) ; X ~ 1,272
Excellent!
Thanks for sharing ❤️
golden ratio always gives me headache
Great sense of humor😀
Thanks for the feedback ❤️
This one is a no brainer. X ='s Elon Musk and the Nature Of Inefficiency In Government. @ 5:56 TH-cam's Algorithm can't determine something so obvious and easy. Welcome too reality. 🙂
Right on😀
Thanks for the feedback ❤️
Just wasted 5 mins watching this video 😂
Don't drag this shit out, it's 3