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A Nice Math Problem | How to solve for X in this Olympiad Problem ?
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- Опубликовано: 22 ноя 2023
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I think it’s a great equation to solve. The question should have stated x is a real number. So there are no domain restrictions for X. However m should always be positive coz it’s the square of x. In this case both 2 and the positive radical value providing two positive and two negative roots. The highest power of x is 8. Since x is real 4 roots are correct.
You reject the second option because you state that x must be a positive integer but then accept the first solution of root 2 - which is not an integer
If x is a positive integer then how sqrt 2 is an integer?
What I really value is the method. The substitution of n divides the expression in to 2 factors each having 4th power of the variable. Then the substitution of m split each of the above factors in to two quadratic factors. Good approach overall.
Solution by insight
Let x^2=t
t^2+rt(t+7)=7
2^2+rt9=7
t^2=2
x=rt2 or -rt2
Is √2 is an integer?
DIFFERENT APPROACH ( find the mistake challenge )
x^4 + √(x^2 + 7) = 7
x^2 = [7 - √(x^2 + 7)]^2
x^2 = 49 + x^2 + 7 -14√(x^2 + 7)
x squrt will cancel out and after furthur solving we will get,
√(x^2 + 7) = 56/14 = 4
x^2 = 16 -7 = 9
x = √9
x = ±3
Challenge -- find mistake
Il faut dire x = ±√9 (et non x = √9 , car √9 = 3 (jamais -3) !!
Sinon (±3)^4 + √((±3)^2 + 7) = 85 (et non 7) ^^
Peux-tu expliquer pourquoi x devrait être un nombre entier ?
m >0 , not x >0
If x>0 then how we will take -√2
Х^Х=7 ?
That's wrong.
Le résultat semble être bon, mais dans son raisonnement, il y a une (ou plusieurs) erreur(s)
Х= Корень из 2.
x=+_V2.