Solving sqrt(1+sqrt(x)) = x-1in Two Ways

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#ChallengingMathProblems #RadicalEquations

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[PS-mhts]

In your first method, you can probably reject (1-√5)/2 sooner (as a candidate value for √x) because √x is expected to be non-negative by convention.

[marcusorban]

I did kind of a mix of both methods:
1. Squaring both sides
2. Sub sqrt(x) = u
Then we get u=0, which we can reject as it isnt in our domain
Then we can factor the rest in u+1 and u^2-u-1 which solutions we have seen in the video

[adipy]

10:25 But √1 is equal to both 1 and -1
Why can't you let x=1 and take the negative answer to the second square root? √(1+√1)=1-1 --> √(1+(-1))=1-1 --> √0=0

[echandler]

substituting y = 1+√x in the original equation gives x = 1+√y

this is an inolution of an increasing function. setting y=x gives x = 1+√x or
x -√x - 1 = 0, a quadratic in √x with + root φ.
x = φ² = φ+1

[madhurkenge]

Finally I solved my first problem on this channel

[pwmiles]

Actually all four roots of the quartic i.e. x=0, x=1, x=(3+/-sqrt(5))/2 fulfil the condition with some choices of sign for the square root terms. For me, negative square roots of reals are permitted, like we use in the quadratic formula.

[konoveldorada]

Finally something that i can solve about Sybermath's problems.

[l.h.]

At 4.39 you have 1 + sqrt(x) = x. This can be solved directly by the quadratic formula by putting x as (sqrt(x))^2, i. e. (sqrt(x))^2 - sqrt(x) - 1 = 0, thus sqrt(x) = (1 + or - sqrt(5))/2, squaring this:
x = (3 + or - sqrt(5))/2. (Or use subst. u = sqrt(x), u^2 = x, 1 +u = u^2, ...)

[CKth]

1. By adding one to both sides we get
x=1+sqrt(1+sqrt(x))
2. Substituting x into the right side we get

3. Doing 2 for many times we get

So we know
x=1+sqrt(x)
3. Solve the above equation

[axbs]

*some madlad just straight up using the cubic quadratic formula*

[armacham]

4:06 who else is disappointed that Syber didn't say "keepin' it real"?

[pk]

If we call the golden ratio phi, the solution to the problem is x = phi^2 . Remember that phi^2=phi+1 . When we substitute x=phi^2 into our original equation we get phi on the left side and phi on the right side . We therefore get that x = phi^2 is a solution . Of course this is in hind site . In reality I solved this problem the hard way before I watched the video . I solved a cubic by Cardano's formula, and then I recognized that the decimal value was phi^2 .

[cicik]

it is the loop equasion. solved by using a cycle
the equasion os √(1+√1+√...) = √x. or shorting the loop √(1+√x) = √x. or

[larzcaetano]

I am SO glad that you do not square both sides at the start. You always try to work around that and that’s something I do too. Squaring both sides might turn out to be extremely bad, not only because you miss a solution, but because it also makes things worse to solve, most of the time.

[benardolivier]

6:14 (1-sqrt(5))/2 is negative and cannot be equal to sqrt(x)

[AntonNazoev]

The both methods are excellent. But there is the third method.

[wuhyltw]

As for the check of the solutions, if you are Japanese, like √ 2, √ 3, √ 7, you remember √ 5 = 2.2360679 in Goro-awase - mnemonic system using the similarity between numbers and words, so (3-√5) / 2 ≒(3-2.236) / 2 = 0.764/2 =0.382<1, (3+√5)/2≒(3+2.236)/2=5.236/2=2.618>1, (3-√5)/2 will be written as unsuitable, and (3+√5)/2 is appropriate.

[samuelbassey]

Wow, I can solve cubic equation and others through your algebraic manipulations, u guys are amazing

[septembrinol]

x^3-4x^2+4x-1 = 0 is hemisymmetric, odd degree. These always have a root of 1.

[tonyhaddad]

Awesome, substitution is so powerful, it make the problem more organized, good job