Germany | A Nice Algebra Problem | Math Olympiad

[@emekaosegbo37]

This is great. I like it.

[@SALogics]

Thanks for liking! ❤

[@LITHICKROSHANMS-gw2lx]

My solution Given equation:- √(6-√(6+x))=x √(6-√(6+x))=√(x)² Now comparing the radical inner term So,it becomes 6-√(6+x)=x² Subtracting '6' on both sides -√(6+x)=x²-6 Squaring on both sides we get 6+x=x⁴-12x²+36 Simply further x⁴-12x²-x+30=0 Now, P₁(x)=x⁴-12x²-x+30....(1) P₁(1)≠0 P₁(2)≠0 P₁(3)=0 So, x=3 is a root x-3 is a factor....(2) (1)÷(2) (x⁴-12x²-x+30)÷(x-3)=(x³+3x²-3x-10) So, P₂(x)=x³+3x²-3x-10....(3) Initial guess x=-2 P₂(-2)=0 So, x=-2 is a root x+2 is a factor....(4) (3)÷(4) (x³+3x²-3x-10.)÷(x+2)=(x²+x-5) P₃(x)=x²+x-5 We can use quadratic formula x=(-1±√21)÷(2) After some checking valid solution of 'x' x=(-1+√21)÷(2)

[@SALogics]

Very nice trick! I really appreciate that ❤

[@kareolaussen819]

To check that ((sqrt(21) -1)/2)^2

[@SALogics]

Very nice! ❤

[@abbasmasum1633]

Good method. But I solved it using Syn, Div. once I formed my 4th degree polynomial, I found the factoring too long and the concept should match the level as factoring is a low level and your example is an upper level of algebra.

[@SALogics]

Very nice! ❤

[@SidneiMV]

√[6 - √(6 + x)] = x x ≥ 0 √(6 + x) = u √(6 - u) = x 6 + x = u² 6 - u = x² u² - x² = u + x (u + x)(u - x) - (u + x) = 0 (u + x)(u - x - 1) = 0 u = - x ∨ u = x + 1 u = -x => 6 + x = x² x² - x - 6 = 0 (x - 3)(x + 2) = 0 x = 3 ∨ x = -2 [ both are invalid ] u = x + 1 => 6 - (x + 1) = x² x² + x - 5 = 0 *x = (-1 + √21)/2*

[@SALogics]

Very nice trick! ❤

[@jackshen9845]

Very good!

[@kareolaussen819]

You can immediately rule out x+u=0, since x>0, u>0. Otherwise very elegant solution❤.

[@bandarusatyanandachary1181]

Why can't we use factor theorum to solve the problem in simplest way.

[@SALogics]

Yes, it can be done so. ❤

[@tomtke7351]

this problem's solution is a bunch of hocus pocus. A lot of fumbling around to factor without any real structure.

[@kareolaussen819]

To solve the equation x^4-12x^2-x+30=0, first search for integer solutions. These must be integer factors (positive or negative) of 30. It is clear that the author knows that two solutions are x=3 and x=-2, and organizes the calculation to factor out (x-3)(x+2). This can be done by long polynomial division, which is essentially what the author does.

[@tttttt58589]

a common task for a Russian school. a class with an in-depth study of mathematics. nothing complicated)

[@SALogics]

Very nice! ❤

[@mrhook1943]

Why not use synthetic division?

[@SALogics]

Nice Suggession! ❤

[@rasulovamahfuza397]

x^4-12x^2-x+30=0 x=-2 16-48+2+30=0

[@Cookielolll]

Plug 2 you get sqart of 6 - sqart sqart of 6+2 = sqart 6 - sqart sqart 8 = sqart 6-2=2

[@Cookielolll]

And x=2

[@SALogics]

Very nice! ❤

[@tomtke7351]

sqrt[6-sqrt(6+x)]=x square {sqrt[6-sqrt(6+x)]=x} 6-sqrt(6+x)=x^2 rearrange x^2 - 6 = -sqrt(6+x) squar{x^2 - 6 = -sqrt(6+x)} 6+x=[x^2-6]^2 ///// this results in an x^4 term yielding there's FOUR solutions. 6+x=x^4-12x^2+36 rearrange x^4 +0x^3-12x^2 -x+30

[@SALogics]

Very nice trick! ❤

[@ДмитрийЗенков-х7о]

Почему 3 не подходит для решения? 6+3=9 корень из 9 = +3, - 3, берём-3 отнимаем от 6=9, корень из 9 = 3, -3 В правой части 3 И получается равно в 50%😂 Квантовая запутанность.

[@SALogics]

3 отклоняется, поскольку квадрат 3 больше 6 и при проверке мы не можем брать -ve значение квадратного корня ❤

[@ДмитрийЗенков-х7о]

@SALogics Откуда вы взяли это ограничение? Проверка осуществляется постановкой на месте Х ответа и при постановке 3 вместо Х, нигде не получается корня из отрицательного числа.

[@ДмитрийЗенков-х7о]

@SALogics По существу это уравнения пересечение прямой и пораболы и прямая пересекает пораболу в 2 х точках.Нарисуйте график и всё станет понятно

[@Qazwsx-m3f]

Схема Горнера

[@SALogics]

Very nice! ❤

[@quynhnguyen3873]

Hằng Đẳng thức hả thầy.

[@stratosleounakis2267]

(6+X)^1/2=Y 6+X=Y^2 (1) (6-Y)^1/2 =X 6-Y =X^2 (2) X>0 , X=0 (2) - (1) : X^2-Y^2+X+Y=0 (X+Y)(X-Y)+(X+Y) =0 (X+Y)(X-Y+1)=0 1. X=-Y (2): 6+X=X^2 X^2-X-6=0 X1=-2 , X2=3 X1=-2

[@SALogics]

Very nice trick! ❤

[@pehsuntung9211]

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