Spain | A Nice Algebra Problem | Math Olympiad

math olympiad
olympiad math
math olympiad question
math olympiad questions
math olympiad problem
math olympiad problems
olympiad math problems
algebra olympiad problems
math
maths
algebra
algebra problem
nice algebra problem
a nice algebra problem
equation
exponential equation
a nice exponential equation
radical equation
a nice radical equation
math tricks
imo
Find the value of x?
How to solve √1 + √1+x = ∛x
In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation √1 + √1+x = ∛x 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 #maths #math #algebra

[@Magn0sC4rLs0n]

Spoiler blocker 🛡️ 🛡️ 🛡️

[@逸園-無毒果園]

x=0 and x=-1 are not solutions x=8 is the only solution

[@SALogics]

You are right! ❤

[@melissajenner8068]

Yes, verification needed.

[@ChrisManton]

True, but the math is correct. How to explain ?

[@CheesedogIsWhat]

@@ChrisManton It's called an extraneous solution. It is like a fake root resulted from a transformed equation which is not directly the same as the original equation. Take ✓x = -1 for example, squaring both sides gives us x = 1, but we know ✓1 is 1, not -1, so 1 is an extraneous solution. So there it is, that's the answer.

[@sohamb4944]

@@CheesedogIsWhat I always wished sqrt(x) was a direct inverse of x^2

[@sy8146]

Thank you for explaining. The solution is only x=8.

[@SALogics]

you are welcome! ❤

[@CharlesChen-el4ot]

(1+x) =( x^(2/3) - 1)^2 1+x = x^(4/3) - 2x^(2/3) + 1 x = x * x ^(1*3) - 2x^(2/3) 1= x ^(1/3)- 2*x^(-1/3) Let y = x^(1/3) 1 = y - 2 /y y^2 - y - 2 =0 (y-2)(y+1)=0 Solution 1 y = 2; x^(1/3)= 2; x=8 Solution 2 y = -1 ; x^(1/3)=-1; x= -1 帶入原式不合 非解 ！

[@SALogics]

Very nice trick! ❤

[@key_board_x]

First solution: x = - 1 √[1 + √(1 + x)] = ³√x → where: x = - 1 √[1 + √(1 - 1)] ? ³√(- 1) √[1 + 0] ? (- 1)^(1/3) √1 ? - 1 1 ≠ - 1 → x = - 1 must be rejected Second solution: x = 0 √[1 + √(1 + x)] = ³√x → where: x = 0 √[1 + √(1 - 0)] ? ³√(0) √[1 + √(1)] ? 0 √[1 + 1] ? 0 √2 ≠ 0 → x = 0 must be rejected

[@SALogics]

Very nice! I really appreciate that ❤

[@taniacsibi6879]

Din cele 3. Sol.doar x=8 verifica ec. Data

[@SALogics]

Da, ai dreptate 8 este singura solutie ❤

[@АндрейПергаев-з4н]

Не верно Если подставьте х=0 0=sqrt 2, Если подставить х=-1 1=-1, что тоже не верно Поэтому корень один х=8, о чём не сказано, значит корня получается три. Ответ полностью не верно

[@SALogics]

большое спасибо! x = 8 включено в решения, но я забыл проверить значения x. Извините за неудобства! ❤