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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

[逸園-無毒果園]

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

[CharlesChen-elot]

(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 ！

[sy]

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

[АндрейПергаев-зн]

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

[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

[ChavoMysterio]

√[1+√(x+1)]=³√x

Domain: x≥0
Let y=³√x, y³=x

√[1+√(y³+1)]=y
y²=1+√(y³+1)
y²-1=√(y³+1)
y⁴-2y²+1=y³+1
y⁴-y³-2y²+1-1=0
y⁴-y³-2y²=0
y²(y²-y-2)=0
y²(y+1)(y-2)=0

y²=0
y=0
³√x=0
x=0 ∅
√2≠0

y+1=0
y=-1
³√y=-1
y=-1 ∅
1≠-1

y-2=0
y=2
³√x=2
x=8 ❤
2=2 ✓

[taniacsibi]

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