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

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Find the value of x?
How to solve x^2 - 6 =√x + 6
In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation x^2 - 6 =√x + 6 in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #mathtricks #maths

[@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 ❤

[@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 ❤

[@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! ❤

[@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!

[@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! ❤