Question

2x^2+√2x^2-4x+1=4x+8

Answer 1

Відповідь: 1

Пояснення: 1

Answer 2

To solve the equation 2x^2 + √2x^2 - 4x + 1 = 4x + 8, we can follow these steps:

1. Combine like terms on both sides of the equation: 2x^2 + √2x^2 - 4x + 1 = 4x + 8 (2 + √2)x^2 - 4x + 1 = 4x + 8

2. Subtract (4x + 8) from both sides of the equation to get all terms on one side: (2 + √2)x^2 - 8x - 7 = 0

3. We now have a quadratic equation in the form ax^2 + bx + c = 0, where a = (2 + √2), b = -8, and c = -7.

4. To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

   Plugging in the values: x = (8 ± √((-8)^2 - 4(2 + √2)(-7))) / (2(2 + √2))

5. Calculate the discriminant and simplify: Discriminant = (-8)^2 - 4(2 + √2)(-7) = 64 + 28(2 + √2) = 64 + 56 + 28√2 = 120 + 28√2

6. Substituting the discriminant back into the quadratic formula: x = (8 ± √(120 + 28√2)) / (2(2 + √2)) x = 4 ± √(30 + 7√2) / (2 + √2)

7. To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator, which is (2 - √2): x = (4 ± √(30 + 7√2) / (2 + √2)) * (2 - √2) / (2 - √2) x = (8 - 4√2 ± 2√(60 + 14√2)) / (2)

So, the solutions for the equation 2x^2 + √2x^2 - 4x + 1 = 4x + 8 are: x = (8 - 4√2 + 2√(60 + 14√2)) / 2 x = (8 - 4√2 - 2√(60 + 14√2)) / 2