Indian Olympiad Mathematics | High school students.

[@BobbyMbock]

From the begining sqrtx=2+sqrt2 and then raise them all to the power of 2. And we are done.

[@dominiquelausseur-michon5027]

complicated and wrong ! I guess you'd better do
√2 + √x = 2
√x = 2 - √2
x = (2 - √2)²
x = 4 - 4√2 + 2
x = 6 - 4√2 ,, that's it ,, 6 + 4√2 is not a solution

[@MaremmaCane80]

I did the same, the funny thing is that he claim to teach math

[@aldoaldoz]

May be an esasier solution is the following.
√2 + √x = 2
means
±√2 ±√x = 2
±√x = 2 ±√2
Since 2 - √2 > 0, we can semplify the above to
+√x = 2 ±√2
Solution 1:
(√x)² = (2 +√2)²
x₁ = 6 + 4√2
[ check: -√2 +√(6 + 4√2) = 2 ]
Solution 2:
(√x)² = (2 -√2)²
x₂ = 6 - 4√2
[ check: +√2 +√(6 - 4√2) = 2 ]

[@Geranb_s_dostavkoi]

Садись, два!!!

[@ericmariaud8237]

Faux, il n'y a qu'une solution.
(√2 + √x)² = 2²
2 + 2√(2x) + x = 4
2√(2x) = 2 - x x > 0 et 2 - x > 0
0 < x < 2
6 + 4√2 > 2 n'est pas une solution
Plus simplement:
√2 + √x = 2 √x = 2 - √2 x = (2 - √2)² x = 6 - 4√2

[@БелАлекс]

Ой, а зачем же так все сложно? sqrt(x)= 2-sqrt(2). Ну и так де как вы возводил в квадрат. Самое главное знак +/- перед корнем не потерять.

[@RealQinnMalloryu4]

{2+2 ➖ }+{x+x ➖ }={4+x^2}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).

[@hassanmorocco9968]

Merci

[@PhilCoolMath]

❤

[@christiann1058]

La tâche aurait été plus simple si 2 passait de l'autre côté de l'égalité. De plus la solution négative ne peut être réponse dans lR

[@PhilCoolMath]

❤

[@sanjiwaranjan8424]

Time killing process

[@Ajaykumar123_rcm]

2

[@thediplomat6165]

3.5

[@PhilCoolMath]

Thanks ❤

[@ΜΙΧΑΗΛΘΕΟΔΩΡΟΠΟΥΛΟΣ-π1ξ]

is is fault
second solution > 10 not verified

[@MahmoudAlJammal]

Sqr2 + Sqr x = 2 =>
Sqr x = 2- Sqr2 =>
X= 4- 2.2.Sqr2 + 2 =>
X= 6-4.Sqr2
Your solution is waaaaaaay toooooooo looooong!!!! , and actually wrong because because x=6+4Sqr2 is wrong buddy!!!

[@christiann1058]

Je veux dire racine de 2

[@PhilCoolMath]

❤