A Hard Radical Math Problem | Can You Solve It?

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A Hard Radical Math Problem | Can You Solve It?
Welcome to "A Hard Radical Math Problem | Can You Solve It?" In this video, we present a challenging radical math problem designed to test your skills and push your mathematical thinking to the limit. Whether you're a math enthusiast, a student preparing for exams, or just love solving tough problems, this video is for you!
In this tutorial, you'll get:
A step-by-step walkthrough of the problem
Key strategies and techniques for solving radical equations
Tips to avoid common mistakes
Insight into advanced problem-solving methods
Think you have what it takes to solve this hard radical math problem? Watch the video, give it a try, and let us know your solution in the comments below. Don't forget to like, subscribe, and hit the notification bell for more challenging math problems and tutorials. Let's tackle this problem together!
Timestamps:
0:00 Introduction
0:40 Graph
1:58 Substitution
2:51 Quintic equation
11:56 Solution
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‪@infyGyan‬

[@NABER-kf6ut]

Thanks

[@mohammedsaysrashid3587]

Thank you Sir...x=0,037 only one real solution

[@sunil.shegaonkar1]

I did not solve this problem but I guessed early on the solution lies between 0 & 0.4.

[@NadiehFan]

First of all, the claim in the video at 0:40 that the graph of y = 1/√x is a hyperbola is _wrong_ because this is actually one branch of the graph of xy² = 1 which is a _cubic_ curve, not a quadratic curve. The graph of y = 1/x or xy = 1 on the other hand _is_ a hyperbola but like all conic sections this is a _quadratic_ curve.
The equation to solve in ℝ is
x² + 5x + 5 = 1/√x
Substituting x = a² and therefore √x = a assuming a > 0 and then multiplying both sides by a we have
(1) a⁵ + 5a³ + 5a = 1
This equation is a special type of quintic equation which can be solved by a hyperbolic substitution. If we substitute
(2) a = 2·sinh(t)
we have
(3) 32·sinh⁵(t) + 40·sinh³(t) + 10·sinh(t) = 1
Dividing both sides by 2 this gives
(4) 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t) = ½
and using the identity
(5) sinh(5t) = 16·sinh⁵(t) + 20·sinh³(t) + 5·sinh(t)
we can write (4) as
(6) sinh(5t) = ½
Since the hyperbolic sine is a periodic function with period 2πi this gives
(7) t = ⅕·arsinh(½) + k·⅖πi, k ∈ ℤ
Substituting (7) in (2) the roots of the equation (1) are
(8) a = 2·sinh(⅕·arsinh(½) + k·⅖πi), k = 0..4
The equation has five roots, four of which are complex. Setting k = 0 gives the only real root, which is
(9) a = 2·sinh(⅕·arsinh(½))
To write this root in algebraic form, we may note that 2·sinh(t) = exp(t) − exp(−t) and arsinh(u) = ln(u + √(u² + 1)). So, we have arsinh(½) = ln(½ + ½√5) and since (½√5 + ½)(½√5 − ½) = 1 we also have −arsinh(½) = −ln(½ + ½√5) = ln(½√5 − ½). Therefore, the real root is
a = 2·sinh(⅕·arsinh(½)) = exp(⅕·arsinh(½)) − exp(−⅕·arsinh(½)) = exp(⅕·ln(½ + ½√5)) − exp(⅕·ln(½√5 − ½)) = ⁵√(½ + ½√5) − ⁵√(½√5 − ½) = ⁵√(½ + ½√5) + ⁵√(½ − ½√5)
and since x = a² this gives
x = (⁵√(½ + ½√5) + ⁵√(½ − ½√5))²
Noting that (½ + ½√5)² = ³⁄₂ + ¹⁄₂√5, (½ − ½√5)² = ³⁄₂ − ¹⁄₂√5 and (½ + ½√5)(½ − ½√5) = −1 the exact solution can be written as
x = −2 + ⁵√(³⁄₂ + ¹⁄₂√5) + ⁵√(³⁄₂ − ¹⁄₂√5)
The numerical value of this solution accurate to seven decimal places is 0.0371649.

[@user-kp2rd5qv8g]

x=0.037