Math Olympiad | A Nice Algebra Problem | A Nice Radical Problem

[@krishnanadityan2017]

Put x+6=u^2 (a)
Subbing,we get x^2-6=u or x^2-u=6. (1)
Also u^2-x=6. (2) (From ran (a))
(1)-(2) givesx^2-u^2-u+x=0
That is (x-u)(x+u+1)=0
Hence either x-u=0 or x+u+1=0
Solving the system involving x and u, we can get values of x more easily.

[@SALogics]

Very nice trick! I really like it ❤

[@НатальяТкаченко-ъ1з]

Очень красивое решение. Спасибо большое

[@SALogics]

Добро пожаловать! ❤

[@АндрейПергаев-з4н]

Пусть 6=у, такая вот замена
х^2-у=sqrt (x+у)
Возводим в квадрат
х^4-2ух^2+у^2=х+у
у^2-(2х^2+1)*у+х^4-х=0
a=1, b=-(2x^2+1) c=x^4-x
Дискриминант d=b^2-4ac=
4x^4+4x^2+1-4x^4-4x=4x^2-4x+1=(2x-1)^2
Корни относительно у
у1=(2х^2+1)+2х-1)/2=х^2+х
у2=(2х^2+1)-(2х-1))/2=х^2-х+1
Поскольку у=6, то получается два простых квадратных уравнения
х^2+х-6=0 и х^2-х-5=0
Решаем, находим корни и проверяем, потому что есть условие х^2>6

[@SALogics]

Очень красивый трюк! Я ценю это ❤

[@oxfordmathseries]

Which app you use to make video? pls reply

[@RyanLewis-Johnson-wq6xs]

x^2-6=Sqrt[x+6] x=3 x=(-1±Sqrt[21])/2=-0.5±0.5Sqrt[21]

[@SALogics]

Very nice! ❤

[@9허공]

domain of x is x^2 ≥ 6 and x ≥ -6 => x ≥ √6 or -6 ≤ x ≤ -√6
you should check the validity of 2nd case.

[@SALogics]

Very nice suggestion! ❤

[@cabasantbab]

X=3

[@SALogics]

Yes, you are right! ❤

[@prollysine]

we get , by faktoring , x^4-12x^2-x+30=0 , (x-3)(x^3+3x^2-3x+30)=0 , x=3 , x^3+3x^2-3x+30=0 , (x+2)(x^2+x-5)=0 , x= -2 ,
x^2+x-5=0 , x=(-1+/-V(1+20))/2 , x= (-1+V21)/2 , (-1-V21)/2 ,
test , 3^2-6=3 , V(3+6)=3 , OK , 2^2-6=-2 , V(-2+6)=+2 , not a solu ,
((-1+V21)/2)^2-6=-1-V21)/2 , V(-1+V21+6)=(1+V21)/2 , not the same , not a solu ,
((-1-V21)/2)^2-6=(-1+V21)/2 , V(-1+V21+6)=(-1+V21)/2 , same , OK , solu , x= 3 , (-1-V21)/2) ,

[@SALogics]

Very nice trick! I appreciate that ❤

[@prollysine]

@@SALogics Thanks , my solution ,
x^4 +/- x^3 - 12x^2 -x + 30 = 0 , multp. *(-3) ,
+1 -3
+3 -9
-3 +9
-10 +30 =0
(x-3)(x^3+3x^2-3x-10)=0 ,

[@SALogics]

@@prollysine Very nice!

[@prollysine]

@@SALogics Thanks!

[@gaiatetuya92]

最後解答が見えない。

[@SALogics]

理由は何ですか? ❤

[@raghvendrasingh1289]

let a= 6
x^4+a^2-2ax^2-x-a = 0
a^2-a(2x^2+1)+x^4-x= 0
discriminant = 4x^4+1+4x^2-4x^4+4x = (2x+1)^2
2a= 2x^2+1+2x+1 , 2x^2+1-2x-1
hence 6 = x^2+x+1 , x^2-x
x^2-x-6=0 , x^2+x-5 =0
rest follows at once.

[@SALogics]

Very nice trick! ❤

[@CharlesChen-el4ot]

x^4 -12x^2 + 36=x+6
x^4-12x^2-x+30=0
x^2*(x+3)(x-3)-
(3x^2+x-30)+0
X^2*(x-3)(x+3)
-(3x-10)(x+3)=0
Solution 1 :
x = -3
x^3-3x^2-3x+10=0
x^3-2x^2-(x^2+3x-10)=0
x^2*(x-2)-(x-2)(x+5)=0
Solution 2 :
x = 2
x^2-x-5=0
x = 1/2(1+/-19i)

[@АндрейПергаев-з4н]

Какие комплексные корни?
В квадратном уравнении
ах^2+bx+c=0
Если а>0 и с

[@SALogics]

Very nice! ❤