Simplifying A Very Radical Expression | Denesting Radicals

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x^x=2^{3x+192}
#ChallengingMathProblems #Radicals
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[mcwulf]

Using the (a + b√5)^2 method, twice, I got the answer.

[sngmn]

Easy solved in mind this first step:
6 + 2*root(5) = (1 + root(5))^2
After adding to 13 + 5*root(5) we got under the main_root 14 + 6*root(5)
We must assume that (14 + 6*root(5)) is an also a perfect square
Ok.
a^2 + b^2 = 14
2*a*b = 6*root(5)
Easy system, a = 3, b = root(5).
Our answer is a root(14 + 6*root(5)) = root((a+b)^2) = a + b = 3 + root(5) = Answer.

Note: root() is a inverse squaring function operator

[vladimirkaplun]

Hassle free nested radicals formula does it automatically .

[張閎茗音樂及數學頻道]

You can make more videos about ordinary differential equations. I have to review because I didn't learn very well.

[goldfing]

O make the ansatz
sqrt(6 + 2*sqrt(5)) = x + sqrt(y)
and square both sides:
6 + 2*sqrt(5) = (x + sqrt(y))^2
= x^2 + 2*x*sqrt(y) + y
= (x^2 + y) + 2*x*sqrt(y)
So I get the equation System
x^2 + y = 6
2*x*sqrt(y) = 2*sqrt(5)
Divide the second equation by 2:
x*sqrt(y) = sqrt(5)
And square it:
x^2 * y = 5
So I have the equation System
x^2 + y = 6
x^2 * y = 5
for the variables x^2 and y. Since I have their sum and product,
I can according to Vieta easily get them by solving the quadratic equation
z^2 - 6z + 5 = 0
(z - 1)(z - 5) = 0
Either x^2 = 1 and y = 5, or x^2 = 5 and y = 1. Since I want integer solutions, I chose the first variant and try
x = 1 and y = 5
Indeed, (1 + sqrt(5))^2 = 1 + 2*1*sqrt(5) + 5 = 6 + 2*sqrt(5).
So I can simplify:
13 + 5*sqrt(5) + sqrt(6 + 2*sqrt(5)) =
13 + 5*sqrt(5) + 1 + sqrt(5) =
14 + 6*sqrt(5).
Now the same procedure for the square root of 14 + 6*sqrt(5):

[isaacdagwom]

Do you have another channel or someone else is impersonating you? It's called sybermaths2