Olympiad Mathematics | Algebra Problem

Olympiad Mathematics | Algebra Problem

[guyhoghton]

That was a nice piece of factorization.
I solved as follows:
With x² = a, y² = b,
x³ + y³ = 3 ... ➀
xy(x + y) = 2 ... ➁
∴ (x + y)³ = x³ + y³ + 3xy(x + y) = 3 + 3.2 = 9
∴ x + y = ³√9 = 3⁽²÷³⁾ ... ➂
and xy = 2/(x + y) = 2/(3⁽²÷³⁾) ... ➃
From Vieta's formula x and y are the roots of:
z² - (x + y)z + xy = 0 ... ➄
⇒ z² - 3⁽²÷³⁾z + 2/(3⁽²÷³⁾) = 0 from ➂ and ➃.
⇒ (3⁽²÷³⁾)z² - 3⁽⁴÷³⁾z + 2 = 0
∴ z = [3⁽⁴÷³⁾ ± √(3⁽⁸÷³⁾ - 8(3⁽²÷³⁾))] / (2.3⁽²÷³⁾)
= [3⁽⁴÷³⁾ ± 3⁽¹÷³⁾√(9 - 8)] / (2.3⁽²÷³⁾)
= [3⁽²÷³⁾ ± (3⁻⁽¹÷³⁾)] / 2
= (3⁻⁽¹÷³⁾)(3 ± 1) / 2
∴ z = 2/(³√3) or z = 1/(³√3).
These are the values of x and y in either order, since they are the roots of ➄.
∴ a, b = 4/(³√9), 1/(³√9) in either order.

[frankxxx]

This is really really hard
As someone else pointed out in the comments, it is much easier to do
Eq.1 + eq. 2= 5
Eq.1 - eq. 2 =1

So

Eq.1 + eq.2 = 5 *(eq.1 - eq. 2)
(a+b)(√a + √b) = 5 ( a-b)(√a -√b)
(a+b)(√a + √b) = 5 ( √a+√b)(√a -√b)^2
a+b = 5 (a+b-2√(ab) )
2a+2b= 5√(ab)
2√(a/b) + 2√ (b/a) = 5

From here you get a=4b OR b= 4a
Then you substitute in eq.1 and get the solution

Nice problem 👍

[satyapalsingh]

Nice method of solving !Keep it up, dear Professor!!!

[КатяРыбакова-шд]

Если умножить второе уравнение на 3 и сложить уравнения, получим куб суммы: (Va+Vb)^3=9. Тогда Va+Vb= 9^1/3 и из второго уравнения Va*Vb=2: (9)^1/3. По обратной теореме Виета получим Va =1/(3)^1/3, Vb=2/(3)^1/3 или наоборот. Ответ: ( 1/(9)^1/3; 4/(9)^1/3) или наоборот.

[levelup]

Add the given equation then once subtract factorise a-b=√a+√b. √a-√b then divide whole equation by √ab form a quadratic in √a/b+√b/a put √a/b =x folve x+1/x=5/2 clearly visible x=2 then back substitute in original equation will get the relationship as a=4b solving this will get you the answer

[mrsahil]

Why these Olympiad problems are easy---

[Metal_dead]

Also you didn't prove that x = -y is not the solution

[Metal_dead]

Слабак. Нормальные люди решают это в уме и без замены