Solving a Golden Radical Equation

[@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.

[@SyberMath]

Nice!

[@italixgaming915]

You still need to check if your candidate is a solution. Here is my method, that uses your idea: 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)

[@shrovitz969]

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.

[@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.

[@gianmarcolettieri6150]

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

[@SyberMath]

Good point!

[@gemeni0]

Положительные то они положительные ежели не комплексные. 😅

[@notananimenerd1333]

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

[@Skank_and_Gutterboy]

That's the way I went with it, too. The roots I got are: -1, 2, (√5-1)/2, and (-√5-1)/2. When checking these values against the original equation, only x=(√5-1)/2 is a good solution.

[@shmuelzehavi4940]

How did you a priori know it should be: (x^2+ax-1)(x^2+bx-2) and not (x^2+ax+1)(x^2+bx+2) ?

[@RexxSchneider]

@@Skank_and_Gutterboy We expect if we solve the quartic to get roots of the equations ±√(2 ± √(x + 2)) = x, so the other roots are the solutions of -√(2 - √(x + 2)) = x for x = -1 +√(2 + √(x + 2)) = x for x = 2 -√(2 + √(x + 2)) = x for x = (-√5-1)/2

[@itsgoodtoplaygames3830]

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.

[@SyberMath]

Not in the real world!

[@vivekkirubakaran6193]

At 4:08, why X should be greater than 0 because it is square root of something? X can be both positive and negative, right? For example X = Square root of 4 means X can be +/-2 (plus or minus 2)

[@SyberMath]

Square root of a real number needs to be non-negative

[@vivekkirubakaran6193]

@@SyberMath Square root of the real number 4 is +/-2. This essentially means square root of a real number can be negative.

[@MrGoofy42]

@@vivekkirubakaran6193 The sqare root of 4 is 2 (it is just defined as a positive number). The solution of x²=4 is +/-2.

[@vivekkirubakaran6193]

@@MrGoofy42 isn't X-Square = 4 same as X = Squareroot of 4?

[@kailashanand5086]

@@MrGoofy42 exactly

[@volodymyr.kushnir]

Did you tried put that answer into input equation ? For me only -1 works, am i doing something wrong ? I used assumptions: 1) x +2 >= 0 thus x >= -2 2) 2 - sqrt (x+2) >= 0 thus 2 >= sqrt(x+2) thus 4 >= x+2 thus 2 >= x So finally X should be in range [-2; 2). So i got 4 answers 2, -1 and same ones as you did. But only -1 works if i put it into original equation.

[@RexxSchneider]

There is a convention (not universally honoured) that the square root sign designates the positive root only. So +√(2 - √(x + 2)) = +√(2 - √(-1 + 2)) for x = -1. That becomes +√(2 - √1) = +√(2 - 1) = +√1 = +1 which is not equal to x when x = -1, so x = -1 is _not_ a solution. The four values you found are the four solutions of ±√(2 ± √(x + 2)) = x as we might expect. But only x = (√5-1)/2 is a solution to the equation in the video.

[@theOman333]

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

[@SyberMath]

Hey, thanks! 🥰

[@emanuellandeholm5657]

(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.

[@SyberMath]

Thank you!

[@firstnamelastname4792]

@@SyberMath it meant (sqrt(5)+1)/2 = 1 - phi = 1/phi

[@firstnamelastname4792]

it really meant (sqrt(5)+1)/2 = 1 + phi = 1/phi !

[@PrimitiveSkillOfHmong]

Good job

[@SyberMath]

Thanks

[@PrimitiveSkillOfHmong]

@@SyberMath welcome,I like your Idea to solve ✌

[@souleymanesylla2548]

Nice thinking

[@SyberMath]

Thanks!

[@robertodiasfb]

Sooooo niiice! Thks

[@SyberMath]

No problem! Glad you like it!

[@michaelpurtell4741]

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

[@SyberMath]

Good!

[@MathZoneKH]

That’s a good solution sir!

[@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.

[@koennako2195]

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 sqrt(a-sqrt(a+sqrt(a-sqrt(a+x)))). 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.

[@guadalajara4848]

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.

[@ibrahimsoubki7587]

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 😉

[@SyberMath]

Thank you, professor! 💖😊

[@paultoutounji3582]

As always I loved the video !

[@SyberMath]

You're the best!

[@mateuszjarek8587]

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

[@alainfontenla7905]

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

[@italixgaming915]

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)

[@ZAHIDHUSSAINBAIG]

As I have calculated that X=-1 is also a solution of this problem because it is satisfying the given equition.

[@SyberMath]

-1 cannot be the square root because x is real

[@theOman333]

is it negative little phi?

[@SyberMath]

Is it? 😁

[@Артьомдругартем]

Можно сделать подстановку x=2cos y 2+2cos y=4cos^2(y/2) 2-2cos(y/2)=4 sin^2(y/4) 2sin(y/4)=2 cos y cos(π/2-y/4)= cos y 5/4y=π/2. y=2/5π

[@SyberMath]

Nice!

[@aashsyed1277]

@@SyberMath real nice!!!!!!

[@AnkitYadav-il2fo]

It was so nice

[@giuseppelucianoferrero8916]

prof. bravo! ma mi domando quale sia la rappresentazione "geometrica" di quel valore di X= 1/𝛗 =(+ 0,618...),ovvero, il reciproco del rapporto aureo! l'equazione (X^2-X-1=0 ); risolve invece in (+𝛗) positivo e ( -1/𝛗) in negativo , il cui prodotto P= -1=( -0,618*1,618) e la cui ∑ = +1=( -0,618+1,618). Inoltre, la Parabola ,che li rappresenta , ha un senso anche nella geometria euclidea cartesiana perché l'unità(= 1) indica il diametro del cerchio in cui è inscritto il triangolo retto di lati b=√ (1/𝛗) ; b= (-1/𝛗) e c=( +b+ 1/𝛗^2) ) i cui valori sono rispettivamente ; a= 0,786..( cateto lungo); b=( - 0,618 ) cateto corto ; c= (0,618.. + 0,382..)=1 ( ipotenusa,diametro) Infine ,faccio notare che il valore negativo del lato corto indica in geometria analitica che la sua pendenza è negativa e ,pertanto si trova nel primo quadrante ,quando gli assi cartesiani passano per l'altezza h=√( 0,618)(0,382)=√ 0,236... ≃=0 ,486.. Mi scuso se mi sono allargato nell'interpretare i risultati ma mi pareva utile per il dibattito scientifico in questione. cordialità. joseph 11/12/21

[@piyushdaga357]

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

[@ashishpradhan9606]

Yup I got also one solution. It's (√5-1)/2. But why have you mentioned about -1 and 2

[@piyushdaga357]

@@ashishpradhan9606 Yup I think so you have tried different method I went through simple approach. I went on squaring this equation until I get a Polynomial of degree 4 . Then I applied factor theorem to factorise the polynomial. From there I got 4 roots of that conic equation. I plugged those roots back in my original equation but only one root actually satisfied the given equation So the rest roots were extraneous solutions

[@ashishpradhan9606]

@@piyushdaga357 Mine approach is also exactly like yours but I avoided the extrenous solutions by restricting the domain of x with inequality. 😁😁😁

[@leif1075]

No negative 1 is also valid.plug it in amd it works out..

[@Qermaq]

@@leif1075 Think about it this way. To get that to work, we must assume the inside radical yields the positive root and the outside a negative. There is ambiguity there. I'd agree that -1 is a valid value we can assign, but I disagree that it's a proper solution. Only 1/phi truly works without ambiguity.

[@ybodoN]

You at 3:35 searching for a name... Me thinking about domino effect...

[@SyberMath]

😁

[@roman_roman_roman]

I am from Siberia, so, SyberMath is good for me :)

[@SyberMath]

That's cool. 😁

[@roman_roman_roman]

@@SyberMath , that"s cold

[@Alians0108]

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

[@SyberMath]

Pretty good!

[@strikerstone]

finally i was able to solve this myself !

[@michaelempeigne3519]

( sqrt 5 - 1 ) / 2 = 1 / phi

[@SyberMath]

That's right!

[@williamemerson7343]

@@SyberMath It is also (phi-1)

[@alainfontenla7905]

Theoretically you are supposed to determine first for which set of x this function exists. You might have a surprise.

[@Ni999]

Not in the mood for real math, strictly my problem and pretty rare. A few minutes of iterative search with a calculator gave me 0.618, not bad. Real math would have been less work lol. Edit - excellent work in the video!

[@SyberMath]

😊

[@sueyibaslanli3519]

The best way is graph, I think

[@SyberMath]

Graphing is almost always a good thing

[@Blaqjaqshellaq]

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

[@kailashanand5086]

after seeing a bunch of mindyourdecision videos i was pretty sure a solution would be root(5)-1/2 haha nice! i didnt think of the second method, i solved it the first way

[@SyberMath]

Excellent!

[@libardouribe883]

Una SUPER estrategia. Felitaciones.👏

[@SyberMath]

Thank you!

[@oguzhanaras2897]

Respect!!!

[@Gezraf]

the golden ratio conjugate

[@gemeni0]

В уме: 2;-1;(±√5-1)/2 Из них положительные: 2 и (√5-1)/2

[@kalyanbasak6494]

Good afternoon, answer sharing X=2& 3 ,sir your good teaching capacity, awesome thanks

[@SyberMath]

Np. Thank you!

[@rocamgreg]

Excelente 👍👍👍

[@SyberMath]

Obrigada!

[@ramlakhanyadav6563]

I think x belong to (0, √2) not (0, 2) please check the domain of x again

[@kurtlichtenstein2325]

Good one!

[@SyberMath]

Thanks for listening!

[@kurtlichtenstein2325]

I wonder, if you knew where you were going, you could aim to prove that ux=1, and u=x+1, and if you knew that, there's only one thing u and x can be.

[@irwandasaputra9315]

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

[@damiennortier8942]

It is phi - 1 or 1/phi

[@nawusayipsunam1643]

Good problem.

[@SyberMath]

Thanks!

[@aagiard]

la réponse est dans l'énoncé du problème : phi = 1 + 1/phi

[@Nikioko]

If they problem is Golden, the answer must be Φ = (1 + √5) / 2. ;-)

[@annoyingbstard9407]

Is it 7?

[@SyberMath]

No

[@annoyingbstard9407]

@@SyberMath It was a good guess though.

[@tonyhaddad1394]

I not sure but i think this is the conjugate of the golden ratio (im right ?)

[@SyberMath]

Sort of but it is also related to Golden Ratio in a different way!

[@angelmendez-rivera351]

It is the multiplicative inverse of the golden ratio.

[@tonyhaddad1394]

@@SyberMath i think angel mendez has get the answer !!!!

[@leonhardeuler5211]

@@tonyhaddad1394 yeah it’s the reciprocal

[@tonyhaddad1394]

@@leonhardeuler5211 thank u

[@nathanruben3372]

if √(x+2) = u, the expression √(2-u) = x is wrong. I did not watch the rest of it after seeing it.

[@SyberMath]

Why?

[@alainfontenla7905]

Exactly. This is not a good way to teach maths. I am sorry to say.

[@ianmathwiz7]

1/φ is the answer.

[@bapibasu2840]

This is Fibonacci's ratio

[@bitsavas]

i want to ask you something mister sybermath..are you a mathimatician ? i suppose yes? you are teaching in a school?

[@SyberMath]

Thanks for asking. I don't know what qualifies for mathematician but I studied math in college. I taught in schools for some time and then stopped. Currently tutoring students one-on-one.

[@bitsavas]

@@SyberMath have a nice day ,where you live in usa i suppose?

[@заводмихельсона]

Есть,по меньшей мере,десяток методов решения этого уравнения.Если автор заинтересуется,могу приобщить его к настоящей математике.

[@mubitajosephmwepu]

Hey inbox me

[@scottlee8270]

Is this a math Olympiad question?

[@SyberMath]

I don't think so but it might as well be

[@compostsfertilizers5471]

Now use the first method... show off !!

[@SyberMath]

What do you mean?

[@compostsfertilizers5471]

@@SyberMath What level is this math ? It's ok i was just joking.

[@SyberMath]

​@@compostsfertilizers5471 Ok, cool! This is probably high school and above, more like competition style

[@compostsfertilizers5471]

@@SyberMath high quality stuff. Please don't stop.

[@SyberMath]

Thank you!

[@karljo8064]

maths is so complicated

[@Biblapghosh]

X=5.3722.....,-0.3722.....,5.70156.....,-0.70156....

[@arekkrolak6320]

"x+2 must be >0 otherwise it is not going to be a real number" - so what's wrong with it not being a real number? what's so special about real numbers? :)

[@SyberMath]

They are REAL! 😁😜

[@tajpa100]

Golden ratio-1

[@danradutz8228]

You have not checked that the solution is

[@falkinable]

The solution is the reciprocal of the golden ratio. 1/φ

[@kubaszabelski2568]

and also φ-1

[@tonyhaddad1394]

Im*

[@Caturiya]

What is the need of the inequalities? Is it not enough to consider the equation 6.54? You confuse harmless students.

[@muiamathsmadeeasy3760]

Which application is this for solving maths

[@АзизханУмархужаев-з8з]

using bezout theorem faster

[@SyberMath]

What is Bezout Theorem?

[@АзизханУмархужаев-з8з]

Sorry. rational root theorem

[@shanmugasundaram9688]

"x" is the inverse of the golden ratio.

[@SyberMath]

Yess

[@박신영-r2n8y]

♡

[@SyberMath]

🥰