Question

4) (x² + 1)² - 3(1-x²) -4 = 0.​

Answer 1

Ответ:

х=1

Объяснение:

(x² + 1)² - 3(1-x²) -4 = 0

$x^{4}$+5$x^{2}$-6=0

• u=$x^{2}$ и $u^{2}$=$x^{4}$

$u^{2}$+5u-6=0

u=1, u= -6

$x^{2} = 1: x=1, x=-1$

x=1

Answer 2

To solve the equation (x² + 1)² - 3(1 - x²) - 4 = 0, follow these steps:

1. Expand the expressions: (x² + 1)² = x⁴ + 2x² + 1 3(1 - x²) = 3 - 3x²

2. Rewrite the equation with the expanded expressions: x⁴ + 2x² + 1 - 3 + 3x² - 4 = 0

3. Combine like terms: x⁴ + 5x² - 6 = 0

Now, you have a quadratic equation in terms of x². To solve it, you can make a substitution. Let's set y = x²:

y² + 5y - 6 = 0

Now, factor the quadratic equation:

(y + 6)(y - 1) = 0

Set each factor equal to zero and solve for y:

1. y + 6 = 0 y = -6

2. y - 1 = 0 y = 1

Now that you have the values of y, you can find the corresponding values of x:

For y = -6: x² = -6 This has no real solutions because the square of a real number cannot be negative.

For y = 1: x² = 1 Take the square root of both sides: x = ±1

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