
Срочно!! Помогите пж A -x^4-2x^2+3=0 B -4x^4+36x^2-80=0 C 3x^4+3x^2-18=0 срочноооо?!!!!!!!!!


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Ответ:
Извини не чем не могу помочь!!!



Quadratic Equations
Let's solve the given quadratic equations step by step.
Equation A: -x^4 - 2x^2 + 3 = 0
To solve this equation, we can substitute a variable to simplify the equation. Let's use u = x^2. By substituting this value, the equation becomes:
-u^2 - 2u + 3 = 0
Now, we have a quadratic equation in terms of u. We can solve it using various methods, such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -1, b = -2, and c = 3. Substituting these values into the quadratic formula, we get:
u = (-(-2) ± √((-2)^2 - 4(-1)(3))) / (2(-1))
Simplifying further:
u = (2 ± √(4 + 12)) / (-2)
u = (2 ± √16) / (-2)
u = (2 ± 4) / (-2)
This gives us two possible values for u:
1. u = (2 + 4) / (-2) = -6 / -2 = 3 2. u = (2 - 4) / (-2) = -2 / -2 = 1
Now, we substitute back u = x^2 to find the values of x:
1. For u = 3, x^2 = 3. Taking the square root of both sides, we get x = ±√3. 2. For u = 1, x^2 = 1. Taking the square root of both sides, we get x = ±1.
So, the solutions to Equation A are x = ±√3 and x = ±1.
Equation B: -4x^4 + 36x^2 - 80 = 0
This equation is a quartic equation, which means it is a polynomial equation of degree four. Solving quartic equations can be more complex than quadratic equations, and there is no general formula like the quadratic formula.
To solve this equation, we can use factoring or numerical methods such as graphing or using a calculator. Let's try factoring first:
By factoring out a common factor, we get:
-4(x^4 - 9x^2 + 20) = 0
Now, we have a quadratic equation inside the parentheses. We can factor it further:
-4(x^2 - 4)(x^2 - 5) = 0
Setting each factor equal to zero, we get:
1. x^2 - 4 = 0: - x^2 = 4 - x = ±√4 - x = ±2


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