Question

X³-3x²+2=0 хеелп плиз​

Answer 1

Ответ: 1 - √3; 1; 1 + √3

Объяснение:

x³ - 3x² + 2 = 0

x³ - x² - 2x² + 2 = 0

x²(x - 1) - 2(x² - 1) = 0

x²(x - 1) - 2(x - 1)(x + 1)  = 0

(x - 1)(x² - 2x - 2) = 0

Произведение равно нулю, когда один из множителей равен нулю, следовательно:

$1)x-1=0\\ \\x_1=1\\ \\ \\ 2)x^2 - 2x - 2 = 0\\ \\ D = 4 - 4\cdot 1 \cdot (-2) = 12 \\ \\ \sqrt{D}=\sqrt{12}=2\sqrt{3}\\ \\x_2=\frac{2-2\sqrt{3} }{2} =1-\sqrt{3}\\ \\x_3=\frac{2+2\sqrt{3} }{2} =1+\sqrt{3}$

Answer 2

Ответ:

х1= 1; х2= 1+√3; х3= 1-√3

Объяснение:

х³-х²-2х²+2=0

х²(х-1)-2(х²-1)=0

х²(х-1)-2(х-1)(х+1)=0

(х-1)(х²-2х-2)=0

х-1=0 или х²-2х-2=0

Если х-1=0, то х1=1

Если х²-2х-2=0, то:

Д=4+4×2=12

х2,3= (2±2√3)/2= 1±√3

Answer 3

To solve the equation x³ - 3x² + 2 = 0, we can use factoring and the zero-product property. Let's break it down step by step:

1. Start with the equation: x³ - 3x² + 2 = 0

2. Look for possible rational roots using the rational root theorem. The possible rational roots are the factors of the constant term (2) divided by the factors of the leading coefficient (1). In this case, the factors of 2 are ±1 and ±2, and the factors of 1 are ±1.

   Possible rational roots: ±1, ±2

3. Test the possible rational roots using synthetic division or substitution to find any roots that satisfy the equation. By testing these values, we find that x = 1 is a root of the equation.

   Perform synthetic division:

   luaCopy code
          1 | 1  -3  2
            |    1 -2
            |-----------
            | 1  -2  0
   

   The result is a remainder of 0, which means x = 1 is a root of the equation.

4. Now, we have factored the equation as (x - 1)(x² - 2) = 0.

5. Set each factor equal to zero and solve for x:

   (x - 1) = 0 --> x = 1

   (x² - 2) = 0

6. For the second factor, x² - 2 = 0, we can solve it by taking the square root of both sides:

   x² = 2

   Taking the square root of both sides:

   x = ±√2

7. Therefore, the solutions to the equation x³ - 3x² + 2 = 0 are: x = 1, x = √2, x = -√2.

I hope this helps! Let me know if you have any further questions.