Question

(x-1)(x+1)(x-3)(x+3)=105

Answer 1

2х-3)(2х+4)=105
2х+1=105
2х=105+1
2х=106
х=106:2
х=53

Answer 2

To solve the equation (x-1)(x+1)(x-3)(x+3) = 105, we can start by expanding the expression on the left side and then rearranging the terms to form a quadratic equation. Here's the step-by-step solution:

Step 1: Expand the expression (x - 1)(x + 1)(x - 3)(x + 3) = 105

Expanding this expression, we get: (x^2 - 1)(x^2 - 9) = 105

Step 2: Simplify the expression (x^2 - 1)(x^2 - 9) = 105

Expanding further, we have: (x^4 - 9x^2 - x^2 + 9) = 105

Combine like terms: x^4 - 10x^2 + 9 = 105

Step 3: Rearrange the equation x^4 - 10x^2 + 9 - 105 = 0

Simplifying further: x^4 - 10x^2 - 96 = 0

Step 4: Solve the quadratic equation

Now, let's substitute a variable to simplify the notation. Let y = x^2. We can rewrite the equation as:

y^2 - 10y - 96 = 0

Using the quadratic formula, we have:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -10, and c = -96. Substituting these values into the quadratic formula:

y = (-(-10) ± √((-10)^2 - 4(1)(-96))) / (2(1))

Simplifying: y = (10 ± √(100 + 384)) / 2 y = (10 ± √484) / 2 y = (10 ± 22) / 2

For the positive root: y = (10 + 22) / 2 y = 32 / 2 y = 16

For the negative root: y = (10 - 22) / 2 y = -12 / 2 y = -6

Step 5: Substitute back and solve for x

Substituting y = 16 into y = x^2, we get: 16 = x^2

Taking the square root of both sides: x = ±√16 x = ±4

Substituting y = -6 into y = x^2, we get: -6 = x^2

This equation has no real solutions since the square of a real number cannot be negative.

Therefore, the solutions to the original equation (x-1)(x+1)(x-3)(x+3) = 105 are: x = -4, 4