A Nice Math Olympiad Exponential Equation | The six solutions.
ฝัง
- เผยแพร่เมื่อ 18 ก.ย. 2024
- Hello my Wonderful family 😍😍😍
Trust you're doing fine 😊
If you like this video on how to solve this nice Math Problem, like and Subscribe to my channel for me exciting videos 💥
Math Olympiad, Algebra, Simplification, Germany Math Olympiad,Math problem-solving, Grade 8 Maths, Grade 9 Maths, Nice Square Root Simplification, Japanese Can you solve, Imo Class 7, Imo Olympiad Class 7, Advanced math concepts,Math competition,Waec, Find the Value of X,Math Olympiad strategies,Challenging Math problems, How to Differentiate, How to Integrate, Thailand junior math Olympiad Problem,
Calculus,Algebraic expressions,Mathematics education,MasterTMathsclass
#matholympiad #algebra#math#simplification #Exponents#vedicmath#viralmathproblem #howto#mathematics#viral #mathematicslesson#calculus #MasterTMathsClass
Interesting, very interesting 👍 👍 👍
Glad you think so
The first answer is obvious X = 6 so I wondered about the other 5 answers. Great video!!
Thanks!
Math for the elite. 6 answers, but only 1 serves a purpose to the world in general. Ocham's razor. This reminds me of something that happened decades ago. During the space race between USA and USSR, it became apparent that a normal pen wouldn't write well in space. USA spent a million bucks developing a pen that would. Cosmonauts just used a pencil. The moral of the story: YES, multiple solutions can be derived, but why bother?
Nice 😊
Será Internet para elite? O una máquina para "ver" el cerebro 🧠 (tomógrafos) para la elite?
I think that (x - 1)^6 = 5^6. means that
(x -- 1) = 5
x = 5+1
x = 6
👍
You miss so much.
Recall that the family of (x - a)^(even power) curves has a shape that is similar to regular old x^2, so there will be two real roots. Since the exponent is 6 then the total number of roots (real + complex) will be 6. That means there are four complex roots in addition to the two real roots. Your second line should read | x - 1 | = 5 ---> x - 1 = 5 and x - 1 = -5, so x = 6 and x = -4 are the real roots. Finding the 4 complex roots requires more work.
Which is only one of six solutions.
(Х - 1)⁶ = 5⁶
Х - 1 = |5|
Х - 1 = 5 Х - 1 = -5
Х = 6 Х = -4
The thing to reinforce here is the concept that was so easily missed. Since this is a 6th degree polynomial, the total number of (real roots plus complex roots) is 6.
Nice
It would be far more useful to viewers to just give an explantion of the 'roots of unity'. Let z = x - 1, and use the polar form z = re^(iθ). the 6 unique solutions to z⁶ = 5⁶ are just r = 5 with iθ = 0, 2πi/6, 4πi/6, 6πi/6, 8πi/6, 10πi/6. Doing all this with the Cartesian form for z is a waste of time, especially since the readers aren't learning something as generally useful as the 'roots of unity'.
Absolutely!!! You are one sane person.
👍
Hvala Vam❤
You're welcome 🎉
Thank you. Greettings from Medellín. Excellent video
You're welcome! Thank you and welcome on board 🎉
خوب معلومه ایکس مساوی با ۶
Nice
Agree with the comment about using z = re^(iθ) - however I did learn (was reminded of something I forgot). The complex graph will be centered around x=1 rather than x=0. Thanks for that.
And it will be a regular hexagon inscribed in a circle with radius 5.
So, the real solutions are −5 + 1 = −4 and 5 + 1 = 6.
The complex solutions are 5/2 · (± 1 ± i√3) + 1.
Because the height of a regular triangle with side length 5 is 5/2 · √3, which is the irreal component, and the real component is of course half the side length 5/2. And the whole is shifted by 1 to the right on the real axis.
👍
2 solutions: 6 and -4. Ready...
👍
7
How 🤗
I x-1I=5, x=6 or - 4
🎉
If one knows the n-th ''Unitroots '' already the solution is immediately clear : x[k] = 1+5*Exp[ i*k*π/3] , k=1,...6.
🎉
I solved this in 2 seconds...
Nice
X = 6
👍
Its crazy. The answer is wrong. X = 6. You only have to applay 6th root.
а где проверка?
why complicate your life, a^x = b^x implies a = b. therefore x - 1 = 5 and x -6
Why are you going to solve it through complex way.
اصلا احتیاج به محاسبه نداره ،توانایی که برابر است پس میشودx-1=5 پس x=6
Solution:
(x-1)^6 = 5^6 |()^(1/6) ⟹
x1/2-1 = ±5 ⟹
x1-1 = +5 |+1 ⟹ x1 = +6 and
x2-1 = -5 |+1 ⟹ x2 = -4
Nice
@@MasterTMathsClass But i have only 2 solutions. You spoke from 6 solutions. How can i get the other 4 solutions?
x = 6 or -4
This is only 1/3 of the solution.
x=6, -4
(x-1)^2=5^2
×^2-2×-24=0
(x-6)(x+4)=0
x=6, -4
🎉
I kinda just solved it like this:
(x-1)^6 = 5^6
(x-1)^6 - 5^6 = 0
a = x-1
a^6 - 5^6 = 0
x-x =0
a = 5
5 = x-1
5+1 = x-1+1
6 = x
(6-1)^6 = 5^6
5^6 = 5^6
x = 6
C'mon - dig deeper, this is so much more than that.
@@baselinesweb There is?
You missed 5/6 of the answer.
@@Nikioko What did I miss
@@Mochi-si1bq You got x₁ = 6. You missed:
x₂ = −4
x₃ = 7/2 + 5√3/2 i
x₄ = 7/2 − 5√3/2 i
x₅ = −3/2 + 5√3/2 i
x₆ = −3/2 − 5√3/2 i
only -4 and 6 are solution, the others are not, since they the Ix -1I is less or greater than 5.
Those are the real solutions
X = 6.. why this much complicated steps. Sq route both side, power 6 will go, u get x-1=5, x=6
Not the only solution. Thanks for your comment
13分鐘...
就寫了這麼普通的解法
X= 6,will be the right answer.
No, this answer is wrong.
x should be equal to 6
x=-4 ; (-5)^6 = 5^6 in integer number set
6
Это только один из корней, а где ещё пять, в том числе четыре комплексных?