Can You Crack This Incredible Radical Challenge?

Can You Crack This Incredible Radical Challenge?

🔍 Brace yourself for an incredible mathematical challenge! 🧠💥 Can you crack this mind-bending radical equation puzzle? In this video, we unravel the complexity, providing insights and strategies to tackle the incredible challenge before you. Join us on this journey of problem-solving and watch as we demystify the extraordinary world of radical equations. Ready to put your math prowess to the test? Let's dive in and crack this incredible code together!

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Topics Covered:
1. Understanding the basics of radical equations.
2. Analyzing the unique properties and substitution of the given equation.
3. Step-by-step approach to solving the radical equation.
4. Tips and tricks for handling tricky radicals like a pro.
5. Algebraic identities and manipulations while solving equations.

Timestamps:
0:00 Introduction
0:30 Substitution
2:13 Algebraic identity
4:50 Sum of cubes
7:25 Sum of fifth power
9:48 Sum of 15th power
10:40 Solution

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🎯 This video is perfect for students, math enthusiasts, or anyone seeking to sharpen their problem-solving skills and gain confidence in dealing with radical equations. 🎓📈

🔔 Challenge yourself and see if you can solve the equation before we do! Hit the like button if you're up for the challenge and remember to subscribe for more exhilarating math content! 🛎️🔔

Don't forget to like, comment, and subscribe to join our math-loving community. Let's get started on this exciting journey together! 🤝🌟

Thanks for Watching!
@infyGyan

[safwanmfarij]

Mathematically the method of solving the equation is totally right
But when i used the calculator to evaluate the answer, and put the x value you found (275807) the answer was imaginary

[PeterMcDaid]

Is anyone interested in a substitution?
Put x=shinA . The radicals become (shinA+coshA)^1/15 and (shinA-coshA)^1/15
Now replacing Shin and Cosh by their exponential forms we have (e^A)^1/15 and (-e^A)^1/15
So the equation readily becomes e^A/15+(-e^A/15)=2
We have a quadratic in e^A/15 which has one positive valued solution =1+2^0.5.
Therefore e^A=(1+2^0.5)^15
So x=shin A =1/2{ ( 1+sqrt2)^15- (1+sqrt2)^-15}
Now 1/(1+sqrt 2) rationalises to (-1+sqrt2)
So shinA=
We have two binomial equations where all the odd terms double up so negate the half and all the even terms cancel out.
So we have

[magnusPurblind]

with (Math)
{
floor((1 + SQRT2)**15 / 2)
}

was my last step - using F12 console to check the arithmetic

[AjaySivaram-byvl]

= ready made the text of team and his own personal information conversation is equal to a specific and then join me at least half of them are from

[slimanebenlala]

But this equation has no real solutions ! ! ! !

[E.h.a.b]

We can approve that a+b = -1/b -1/a = -(a+b)/ab so ab=-1 [2]
substitute in [1] we get a(2-a) = -1
a^2- 2 a -1 =0 then solve for [a] we get
a = 1 + Root(2)
Now we can get a^2 then a^4 then a^8 then a^16 = 665857+ 470832 Root(2)
a^15 = a^16 / a = (665857+ 470832 Root(2))/(1+Root(2))
a^15 = 275807+ 195025 * root(2)
Now solve for x we get
a^15 = x + Root(x^2+1)
a^15 -x = Root(x^2+1)
a^30 + x^2 - 2 a^15 x = x^2 + 1
a^30 -1 = 2 a^15 x
x = (a^30 -1)/(2 a^15) = (a^15-1)(a^15+1)/(2 * a^15) = 275807