A nice Math Olympiad Radical Problem | International Olympiad Algebra Solution. Know this Trick

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In this Math Olympiad Algebra Equation, you'll learn tips and tricks of solving International Math Olympiad exams fast. #IMO #matholympiad #algebra #radicalequations #simplify

[@KalyanVajapeyajula]

Case 4 should be X = -3.

[@superacademy247]

Yes. It is an error!

[@rob876]

let √(12 + x) = y then √(12 + y) = x so 12 + x = y^2 and 12 + y = x^2 so (x - y) + (x^2 - y^2) = 0 (x - y)(1 + x + y) = 0 x = y or x = -y - 1 if x = y, x^2 - x - 12 = 0 => (x - 4)(x + 3) = 0 => x = 4 or x = -3 (extraneous) if x = -y - 1 12 - x - 1 = x^2 => x^2 + x - 11 = 0 => x = (-1 ± √45)/2 (extraneous) so x = 4

[@prime423]

Simply let 12+x =a squared. We get a fourth degree equation whose numerical roots are easily found.Using synthetic division the other two irrational roots are obtained.Sometimes the obvious is better than other methods.

[@richardleveson6467]

It's a clever solution - avoiding having to solve a fourth degree equation in x. The strategy depended on the 4x^4s cancelling but, personally, I would never have predicted that! I gave up. Thank you for showing this ingenious method - very entertaining!

[@superacademy247]

Good point!

[@sy8146]

Thank you for explaining. (We don't have to spend much time for checking.) I guess x is a real number. If so, from the given equation, x≧0. Then, (x=)√12(+√(12+x)) > √12 > 3 . Therefore, except for Case 3, we can reject the cases soon without checking calculation. (Case 1: x

[@superacademy247]

Thanks for your support

[@ManjulaMathew-wb3zn]

If you figured out the domain of x first x>=2 sqrt3 and the only solution satisfying this condition is 4.

[@superacademy247]

Thanks 👍💯 for your input

[@leoliu7243]

√(12+x)=x

[@edsonstarteri6313]

2.854 is also a solution assuming that sqrt(12+2.854)

[@edsonstarteri6313]

The same applies to -3.

[@abdulalhazred6328]

in the real world a square root is never negative

[@aperinich]

in mathematics, even numbered roots can be positive or negative, and given that there's no real-world application stated here, your comment is not worthwhile@@abdulalhazred6328

[@anestismoutafidis4575]

x=4

[@souzasilva5471]

There is an error: x = -3 and not 3.

[@superacademy247]

Thanks for your support

[@jamesyeh1524]

Guessing X = 4!

[@aperinich]

i got it from looking at the original statement. it's the only thing that made sense for an example....

[@guyhoghton399]

Let f(x) = √(12 + x) The equation can be written: _f(f(x)) = x_ ... ① Let _x₁_ be a root of ① i.e. _f(f(x₁)) = x₁_ Suppose that _f(x₁) ≥ x₁_ ... ② ⇒ _f(f(x₁)) ≥ f(x₁)_ because _f(x)_ is a strictly increasing function ⇒ _x₁ ≥ f(x₁)_ which from ② ⇒ _f(x₁) = x₁_ In the same way if _f(x₁) ≤ x₁_ then also _f(x₁) = x₁_ So any root of _f(f(x)) = x_ is also a root of _f(x) = x_ The converse is also true because if _f(x) = x_ then _f(f(x)) = f(x) = x_ Therefore we only need to solve the equation _f(x) = x_ i.e. _√(12 + x) = x_ ... ③ ⇒ _12 + x = x²_ ⇒ _x² - x - 12 = 0_ ⇒ _(x + 3)(x - 4) = 0_ Both sides of ③ must be positive, so _x > 0_ ∴ _x = 4_ is the only solution.

[@superacademy247]

Nice 👍💯 method and analytics

[@christianthomas9863]

The way I did it. faster.

[@PavelYakovleff]

X>0

[@GZTrik]

Μετέφερες τις τελευταίες 2 λύσεις από 4 ,-3 σε 4 , 3. Λάθος.

[@Prypak]

x=4 is trivial