Solving a Golden Radical Equation

This video is about solving a radical equation in a not-so-unusual way
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EXPLORE:

[IlanAmity]

You can continue with the polynomial equation x^4-4x^2-x+2=0 by dividing it (long division) with (x+1)(x-2)=x^2-x-2 and get x^2+x-1=0 which yields the two other solutions x=(-1±√5)/2 of which only the positive root (φ) is the right one.

[shrovitz]

You can easily solve the quartic equation here, just factorise it as following
x^4-4x^2-x^2+2=0
write it as
(x^4-x^2)+(-3x^2-x+2)=0
It will be factorised as
x^2(x^2-1)+(2-3x)(x+1)=0
After further factorisation, it will become
(x+1)(x^2(x-1)+2-3x)=0
(x+1)(x^3-x^2-3x+2)=0
We get our first solution, which is -1 but the problem is that it doesn’t satisfy the equation, so it’s wrong. What it means is that (x^3-x^2-3x+2) is equal to 0.
Now, factorise this polynomial, it will factorise as following
x^3-2x^2+(x^2-3x+2)=0
Factorise the quadratic side
x^2(x-2)+(x-1)(x-2)=0
(x-2)(x^2+x-1)=0
2 doesn’t satisfy the equation as well, so it’s not 2?
(x^2+x-1) should be equal to 0.
Now we all know this famous quadratic, the solution comes out to be the golden ratio.

[gianmarcolettieri]

Great video. I just want to point out that you could just have devided by u+x. That's because u=sqrt(x+2), so u=>0 and 0<x<2, so x and u are both always positive. The sum of two positive number is always different from zero as long as at least one of them is different from zero.

[itsgoodtoplaygames]

Great video but -1 is actually a solution as the square root of a number can be either positive or negative resulting in √1=1 or -1 which is the initial value of x.

[RexxSchneider]

Without wishing to be rude, at 2:20 you have the equation x^4 - 4x^2 - x + 2 and suggest that solving it is long-winded, but you then spend more than 10 minutes using a substitution to solve it.
Yet the rational root theorem suggests trying ±1 and ±2, and that quickly shows x = -1 and x = 2 are roots of the quartic. Polynomial division by (x+1)(x-2) gives x^2 + x - 1 = 0 and the solutions of that are (-1 ±√5)/2 which are -ϕ and 1/ϕ, where ϕ = (√5+1)/2, the golden ratio.
Since √(something) = x we know that x is not negative, so we immediately discard x = -1 and x = -ϕ. We check that x = 2 is not a solution, then find that x = 1/ϕ is a solution.

[notananimenerd]

I used your previous factorizing method as
(x²+ax-1)(x²+bx-2) and ended with all the 4 solutions though 1 is correct in the set of real numbers

[theOman]

And cool vid! I'm surprised your channel has the amount of views it has... I expect it to grow, this is quality content!

[italixgaming]

My solution (of course way faster - plus the fact that yours is not complete since you don't prove that there is a solution so you need to test your candidate):
First of all we see that the function f: x---->sqrt(2-sqrt(2+x))-x is defined for x between -2 and 2. We already know, looking at the equation, that the solution must be positive.
The function is continuous, and since f(0)>0 and f(2)<0, the theorem of intermediate values tells us that the value 0 is reached by f. And since the function is strictly decreasing, we know that we have a unique solution s with 0<s<2.
For 0<x<2, our equation is equivalent to: x²-2=sqrt(x+2)
Since the RHS is always positive we must have x>sqrt(2) (sqrt(2) itself is not a solution)
So our original problem is equivalent to: (x²-2)²=x+2 where sqrt(2)<x<2
We're running into a quartic equation and things may become a bit spicy if we can't find any tricks. A first trick that works is to substract (x+2)² to both sides of the equation and we can simplify by x+1 in the end of the process but we would still have to deal with a cubic equation and there is something much quicker here: just solve (x²-2)²=x+2 in Z.
The first square we can have in the LHS is 0, that would force x=-2 but it doesn't work (we would get 4 in the LHS).
The following square is 1, it would force x=-1 and it works.
The following square is 4, it would force x=2 and it works.
Okay, we have two integer solutions, that's enough to finish the problem. Now we just need develop the expression (x²-2)²-(x+2) and to divide this polynomial by (x+1)(x-2)=x²-x-2
We have:
We must have a=1 and c=-1 and from the term in "x" we have: -2b-c=-1 => b=1.
So our problem becomes (x²-x-2)(x²+x-1)=0 where sqrt(2)<x<2 and even x²+x-1 where sqrt(2)<x<2 since the two solutions of x²-x-2=0 are not in our interval.
There is only one positive solution to x²+x-1=0, which is (-1+sqrt(5))/2.
We don't even need to check if it's in our interval since we already know from the beginning that there is one solution.

[ieatgarbage]

2-sqrt(x+2)=x^2
of course this’ll give us some extraneous solution, probably
2-x^2=sqrt(x+2)
4-2x^2+x^4=x+2
aye a real guess and check problem
x^4-4x^2-x+2=0
x=-1
(x+1)(x^3-x^2-3x+2)=0
x=2
(x+1)(x-2)(x^2+x-1)=0
x=[-1+sqrt(5)]/2, [-1-sqrt(5)]/2
The three extraneous solutions were -1, 2, [-1-sqrt(5)]/2. The actual solution was [-1+sqrt(5)]/2.

[emanuellandeholm]

(sqrt(5)-1)/2 = 1 - phi = 1/phi :) Nice video. I tried the sub u = 2 -x which kind of works, but your solution is way cleaner.

[rajeshbuya]

Why is x=2 rejected? In that case, u=√(x+2) could equal +2 or -2, if latter, then √(2-u) would in fact really equate to x=-2, again.

[koennako]

We can use a formual here. For a infinitely nesting radical that goes (where a is a constant) sqrt(a-sqrt(a+x))=x. Like in blackpenredpens video about tejas' challenge problem, we can rewrite this as We can actually keep going forever and ever, so "dot dot dot". How does this have any relevance to this problem. Well, we can just move the 2 in front of the x in the last radical. Then we can see that the pattern is the same that I have described where the signs of the square root change from -, +, -, +, -, +, -, + and dot dot dot. Next, the formula. The formula is (sqrt(4a-3)-1)/2. If we plug in 2 for the variable a, then we get (sqrt(4(2)-3)-1)/2 = (sqrt(5)-1)/2 There are some math videos youtube that describe how to derive this formula. Yes, it is an obscure formula but, if you are trying to learn how to solve problems like these or just challenge problems in general, then you can memorize this formula for an instant answer.

[ibrahimsoubki]

Hi, I'm a math teacher and has just discovered your channel. I enjoyed this video and I'm looking forward to see the others. + 1 subscriber 😉

[michaelpurtell]

I did it the hard way by solving for U which does give the golden ratio as a solution and then converting that back to X

[guadalajara]

After two minutes, you can solve the equation obviously by factorising : X^4 - 4X^2 - X + 2
= X^2 (X^2 - 4) - (X - 2) = X^2 (X - 2) (X + 2) - (X - 2)
= (X - 2) ( X^2 (X+ 2) - 1) = (X-2) (X^3 + 2 X^2 - 1)
and you easily see that the second polynomial is equal to zero when X = -1
so it is divisible by (X+1)
etc.

[alainfontenla]

Apply your method to sqr(2-sqr(x-2))=x. Never start calculating without checking for which x what you are writing exists.

[mateuszjarek]

After finding the quadratic, we can Just check whether divisors of 2 are sols. We easily find them and then use Horner's schema

[Blaqjaqshellaq]

X is the inverse of the golden ratio, as well as the golden ratio minus 1.

[Alians]

My method is pretty generic:

√(2-√(x+2)) = x (2 > x > 0 is implied)
x^2 = 2-√(x+2)
x^2-2 = √(x+2)
x^4-4x^2+4 = x+2
x^4-4x^2-x+2 = 0

x^2(x^2-4) - (x+2) = 0
x^2(x+2)(x-2) - (x+2) =0

(x-2)(x^3+2x^2-1) = (x-2)(x^3+x^2+x^2-1) = (x-2)(x+1)(x^2+x-1) = 0

(x-2)(x+1) are false factors and x^2+x-1 has one negative root. Only valid root is x = (√5-1)/2

[piyushdaga]

(√5/2 ) -(1/2) is the only solution
-1, 2, (-√5/2 - 1/2) are the extraneous solutions