2root(x)-root^4(x)=1

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вопрос

To solve the equation 2√(x) - √(x)^4 = 1, let's first simplify the expression by expanding the term √(x)^4.

Remember that √(x)^4 is the same as (√(x)^2)^2, which is equal to x^2.

So, the equation becomes:

2√(x) - x^2 = 1

Now, let's isolate the square root term by moving x^2 to the other side of the equation:

2√(x) = x^2 + 1

Next, square both sides of the equation to eliminate the square root:

(2√(x))^2 = (x^2 + 1)^2
4x = x^4 + 2x^2 + 1

Now, let's rearrange this equation into a quadratic equation:

x^4 + 2x^2 + 1 - 4x = 0
x^4 + 2x^2 - 4x + 1 = 0

This is a quadratic equation in terms of x^2. Let u = x^2, then the equation becomes:

u^2 + 2u - 4u + 1 = 0
u^2 - 2u + 1 = 0
(u - 1)^2 = 0
u = 1

Now, substitute back for u:

x^2 = 1
x = ±1

Therefore, the solutions to the equation 2√(x) - √(x)^4 = 1 are x = 1 and x = -1.