Question

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

Answer 1

Данное уравнение является биквадратным относительно 2x-1 ⇒ пусть

(2x-1)² = t (t ≥0), тогда уравнение примет вид: t² + 2t - 2 = 0.

D = 2² - 4·1·(-2) = 4 + 8 = 12, √D = √12 = 2√3, t₁ = (-2 + 2√3)/2 = √3 - 1 > 0,

t₂ = (-2 - 2√3)/2 = - √3 - 1 < 0, что не удовлетворяет условию t ≥0. Поэтому второй корень мы отбрасываем, оставляя t = √3 - 1. Далее делаем обратную замену: (2x-1)² = t ⇒ (2x-1)² = √3 - 1 ⇒ (см. рисунок)

Answer 2

To solve the equation (2x - 1)^4 + (2x - 1)^2 - 2 = 0, you can use a substitution to make it easier to work with. Let's set y = (2x - 1)^2. Then the equation becomes:

y^2 + y - 2 = 0

Now, you can solve this quadratic equation for y:

y^2 + y - 2 = 0

To factor or use the quadratic formula, you can write the equation as:

(y + 2)(y - 1) = 0

Now, you have two possible solutions for y:

1. y + 2 = 0 => y = -2
2. y - 1 = 0 => y = 1

Now, you need to substitute back to find the values of x:

For y = (2x - 1)^2:

1. (2x - 1)^2 = -2

2. (2x - 1)^2 = 1

3. (2x - 1)^2 = -2 is not possible in the real numbers because the square of a real number is always non-negative. Therefore, there are no real solutions for this case.

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

Now, you have two cases to consider for this equation:

Case 1: 2x - 1 = 1 2x = 2 x = 1

Case 2: 2x - 1 = -1 2x = 0 x = 0

So, the real solutions to the original equation are x = 0 and x = 1.