Philippine mathematical Olympiad. An intriguing Olympiad math problem 

I think you missed out two more roots.

Clear the fraction and let a=x^2-6 and b=x. You'll get a much simpler problem. The solution will be a nested root but will de-nest nicely.

x^2=t >0 and the equation becomes (t-6)^2=2*t so t^2-14t+36=0 and t=(14+ -2* sqrt13)/2. 1st way of writing is x^2=7+ -sqrt13 > 0 BOTH ACCEPTED, so we have FOUR SOLUTIONS x=+sqrt(7+sqrt13) or -sqrt(7+sqrt13) and +sqrt(7-sqrt13) or -sqrt(7-sqrt13). 2nd way is isolating the square root of square root as foliow. 14+ - 2*sqrt13=13+1+ - 2*sqrt13*1=(sqrt13)^2 +1^2 + - 2*sqrt13*1= (sqrt13+ -1)^2 so x=+ - sqrt{(sqrt13+ -1)^2}/sqrt2=+ -Isqrt13+ -1I/sqrt2=+ -(sqrt13+ -1)/sqrt2= + - (sqrt26+ -sqrt2)/2 .Again 4 Solutions.**sqrt(x^2)=IxI=+ -x so sqrt((sqrt13-1)^2)=I sqrt13-1I=sqrt13-1 because sqrt13-1>0**

Curious why you decided to solve for y instead of x when given the option? Feel like this was a lot of unnecessary work

Yes. He did. The other roots are x≈±1.84 thus x≈±3.26 😅

Let t = x/6 - 1/x (t - 1/6)^2 + (t + 1/6)^2 = 1/6 2t^2 + 2/36 = 1/6 18t^2 = 1 t = x/6 - 1/x => x^2 - 6tx - 6 = 0 x = 3t +- sqrt(36t^2 + 24)/2 x = 3t +- sqrt(26)/2 x = 3t +- sqrt(13)/sqrt(2) 3t = +- 3 / sqrt(18) = +- 1 / sqrt(2) x_1234 = [+-1 +- sqrt(13)] / sqrt(2)

2[(x² - 6)² + x²] = 6x² (x² - 6)² + x² = 3x² x⁴ - 14x² + 36 = 0 x² = (14 ± 2√13)/2 x² = 7 ± √13 *x = ± √(7 ± √13)*

Делим на х и умножить на 36 (х-6/х-1)^2+(х-6/х+1)^2=6 Легко видно что а=х-6/х Тогда (а+1)^2+(а-1)^2=6 а^2+1=3 а=sqrt 2 или а= - sqrt 2 Дальнейшее просто Два простых квадратных уравнений х-6/х-sqrt2=0 или х-6/х+sqrt 2=0 Решение которых не составит труда Решение от силы 3 - 4 минуты, но не 10 как на видео. Если это часть теста, то вы потеряли 6 минут просто так

I disagree with your solution, sir. I have X = ±√(7+√13) and X = ±√(7-√13). They're supposed to be four roots

No substitution (x²-x-6)²+(x²+x-6)²=6x² (x²+x-6-x²+x+6)²+2(x²+x-6)(x²-x-6)=6x² 4x²+2(x²-4)(x²-9)-6x²=0 ÷2 x⁴-14x²+36=0 ⇒(x²-7)²=13 x=±sqr(7±sqr13) By the way your voice is similar to affrican people You from affrica? My greeting to you

iki kök eksik bulmuşsunuz.

2* (x^2 - 6) ^2 + 2 * (x )^2 = 6 x^2 ( x^2 - 6)^2 = 2 x^2 ( x^2 - √2 x - 6) ( x^2 + √2 x - 6) = 0 (( x - 1/√2) ^2 - 13/2) * (( x + 1/√2) ^2 - 13/2) = 0 x = (1 + √(13)) /√2, (1-√(13)) /√2, -(1 + √(13)) /√2, (-1+√(13)) /√2,

Philippine mathematical Olympiad: [(x² - x - 6)/6x]² + [(x² + x - 6)/6x]² = 1/6 x ≠ 0, [(x² - 6) - x]² + [(x² - 6) + x]² = (6x)²(1/6) = 6x², 2[(x² - 6)² + x²] = 6x² (x² - 6)² - 2x² = 0, (x² - 6)² - [(√2)x]² = 0, [x² - 6 - (√2)x][x² - 6 + (√2)x] = 0 x² - (√2)x - 6 = 0 or x² + (√2)x - 6 = 0 x = (√2 ± √26)/2 or x = (- √2 ± √26)/2 Answer check: [(x² - x - 6)/6x]² + [(x² + x - 6)/6x]² = 1/6 x = (√2 ± √26)/2: x² - (√2)x - 6 = 0, x² - 6 = (√2)x [(√2 - 1)² + (√2 + 1)²]/36 = (3 + 3)/36 = 1/6; Confirmed x = (- √2 ± √26)/2: x² + (√2)x - 6 = 0, x² - 6 = - (√2)x [(- √2 - 1)² + (- √2 + 1)²]/36 = [(√2 + 1)² + (√2 - 1)²]/36 = 1/6; Confirmed Final answer: x = (√2 + √26)/2; (√2 - √26)/2; (- √2 - √26)/2 or (- √2 - √26)/2

Horrible video. You would have obtained a polynomial of degree 4 and you only found 2 solutions. You cannot disregard the plus or minus sign in this case.