Russia | A Nice Algebra Problem | Math Olympiad
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- Опубликовано: 5 июл 2024
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I HAVE A easier way to solve this
To solve the equation x+√(x²+√x³)+1=1, we can start by simplifying the square root terms.
Let's first simplify √x³:
√x³ = x^(3/2)
Now, let's substitute this back into the equation:
x + √(x² + x^(3/2)) + 1 = 1
Next, we can simplify the square root inside the parentheses:
√(x² + x^(3/2)) = √(x^2 + x^(3/2))
We can rewrite x^2 as x^(4/2) to simplify the square root:
√(x^2 + x^(3/2)) = √(x^(4/2) + x^(3/2))
Now, we have a common base x in the terms inside the square root, so we can combine them:
√(x^(4/2) + x^(3/2)) = √(x^(4/2)*(1 + x^(1/2)))
√(x^(4/2)*(1 + x^(1/2))) = √(x^2*(1 + x^(1/2)))
Now, substituting this back into the equation, we get:
x + √(x^2*(1 + x^(1/2))) + 1 = 1
This simplification shows that the original equation is not solvable as presented, as the expression x^2*(1 + x^(1/2)) within the square root cannot lead to a solution with x.
√( x² + √(x³ + 1) ) = 1 - x
x² + √(x³ + 1) = x² - 2x + 1
√(x³ + 1) = 1 - 2x
x³ + 1 = 4x² - 4x + 1
x³ - 4x² + 4x = 0
x( x² - 4x + 4 ) = 0
*x = 0*
x² - 4x + 4 = 0
x = 4/2 => x = 2 (not valid)
If 2 is not an answer, why does it appear?? Would be any mistake?!?!
No, it's not
In any equation, if you square both side and solve then you can get some extra solution of that equation. So, you need to check the solution by putting it in original equation.
x=2 "works" if you take the negative sign with both square roots
2 - sqrt(2^2 - sqrt(2^3+1))
= 2 - sqrt(4 - 3)
= 1
Ở hàng thứ 3: (/x^2 +/3 ^3+1)^2 = (1-x)^2....=> (1-x) >= 0... => ( x
Проверять вовсе не нужно было
Когда возводили в квадрат на 1:48, обязательно надо было написать что 1-х>0, то есть х0,
То есть х