A Cubic System Solved in Two Ways

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[@GirishManjunathMusic]

a³ + b³ = 2√5 - ①
a²b + ab² = √5 - ②
Multiplying both sides of ② by 3:
3ab(a + b) = 3√5 - ③
Adding ① and ③:
a³ + b³ + 3ab(a + b) = 3√5 + 2√5 = 5√5
(a + b)³ = 5√5
a + b = √5 - ④
From ②:
ab(a + b) = √5
Substituting in ④:
ab√5 = √5
ab = 1
b = 1/a - ⑤
From ④ and ⑤:
a + 1/a = √5
As a ≠ 0 by definiton:
a² + 1 = √5a
a² - √5a + 1 = 0
Using the quadratic fromula:
a = √5/2 ± 1/2
As the equation set is symmetric, b is the conjugate value.
(a,b) = (½√5 ± ½)

[@carly09et]

@@WahranRai :( b is the conjugate. You need to look up the FULL definition of conjugate. Conjugates are defined relative to the field/ring. X.Y == I makes X conjugate of Y, the operator [.] and identity {l} can be quite general.

[@carly09et]

conjugate is not only the complex conjugate... :(,

[@macelomendes]

👏👏👏👏👏👏hugs from Brazil!!

[@mathswan1607]

a+b=sqrt(5) and ab=1
a and b are the roots of t^2-sqrt(5)t +1=0

[@user-ld5cf5bh4t]

Nice method .Thank you .I solved it the second way .Bravo

[@SyberMath]

Great 👍

[@owlsmath]

Nice problem! Thanks.

[@SyberMath]

No problem!

[@alnitaka]

At 2:18 you say a+b=s and ab=p. At 4:57 you say that s=sqrt(5) and p = 1. This means s and p are roots of x^2-sqrt(5)x + 1. So my idea is to multiply this by x^2+sqrt(5)x+1 to get x^4-3x^2+1, where we can use the quartic formula. The resolvent cubic is y^3-6y^2+5 = 0 which has roots 0, 1, and 5. So a root of the quartic is (1/2)*(sqrt(0)+sqrt(1)+sqrt(5)) = (sqrt(5)+1)/2, the golden ratio. Another root is (1/2)*(-sqrt(0) - sqrt(1)+sqrt(5)) = (sqrt(5)-1)/2. Then continue like you say past 4:57.

[@imonkalyanbarua]

I am thoroughly enjoying your channel and watching as many videos as possible. You Sir are amazing and doing a yeoman service to the community of maths enthusiasts! Bravo! 😇🙏

[@SyberMath]

Awesome, thank you! 💖

[@imonkalyanbarua]

@@SyberMath 😇🙏

[@SuperYoonHo]

Awesome!!!

[@SyberMath]

Thanks!!

[@pranavamali05]

👋

[@jamescollis7650]

Nice system. Would there be complex solutions also?

[@angelmendez-rivera351]

Yes, there would be.

[@mahmoudalbahar1641]

Many thanks to you .
this is very much (cool+clever).

[@SyberMath]

Np.
Thank you! 😍

[@Jha-s-kitchen]

wow!

[@MathAcademy2021]

hi great job🌺

[@SyberMath]

Thank you 🤗

[@zana4464]

سلام استاد عزیز. لطف کنید کانالهایی مشابه کانال شما برای المپیاد فیزیک معرفی کنید. خیلی خیلی سپاسگزارم

[@stevenlitvintchouk3131]

I used the second method. Easy!

[@devondevon4366]

Answer when a=1.61802 b= 0.61802, and when b=1.61802 a=0.61802
This is the Golden ratio
Since (a+b)^3 expands to a^3 + b^3 + 3 (a^2b + b^2 a) then
(a+b)^3 = 2sqrt 5 + 3( sqrt 5) substituting the value of the two equations
= 2 sqrt 5 + 3 sqrt 5
= 5 sqrt 5
=(5^1/2)( 5^1); hence
a + b = (5^1/6)(5^3) cube both sides
= (5^1/6)(5^3)
=5^3/6
a+b = sqrt 5
ab^2 + a^2b = sqrt 5 (given)
ab(a+b) = sqrt 5 factor out 'ab'
ab (sqrt 5)= sqrt 5
ab = sqrt 5/ sqrt 5
ab = 1
So I will use 'ab =1" and 'a+b = sqrt 5" the find the value of "a" and "b"
a = 1/b hence
1/b + b = sqrt 5
1 + b^2 = (sqrt 5)b multiply both sides by 'b"
b^2 - sqrt 5b + 1 =0 using the quadratic formulae gives
b = 1.61802 and b= 0.61802 ANSWER
a= 1/1.61802 = 0.61802 ANSWER
a= 1/0.61802 = 1.61802 ANSWER

[@phxealyrics1004]

Hello sir 😁💘

[@jimmykitty]

Awesome!!!

[@SyberMath]

Thanks!!

[@jimmykitty]

@@SyberMath welcome Sir 💖