Question

3x²(x-1)(x+1)-10x²+4=0​

Answer 1

3x²(x-1)(x+1)-10x²+4=0,

3x²(x²-1)-10x²+4=0,

3x⁴-3x²-10x²+4=0,

3x⁴-13x²+4=0,

Пускай x²=t, t>0; тогда уравнения примет вид:

3t²-13t+4=0,

D=(-13)²-4•3•4=169-48=121,

t=(13+√121)/6=(13+11)/6=24/6=4.

t=(13-√121)/6=(13-11)/6=2/6=1/3.

Сделаем обратную замену, t=x²:

x²=4,

x²=1/3.

Значит x=±2, x=±√3/3.

Answer 2

To solve the equation 3x²(x-1)(x+1) - 10x² + 4 = 0, you can start by simplifying the expression and then factor it further. Here's how you can do it:

1. Distribute the 3x² to the terms inside the first set of parentheses: 3x⁴ - 3x²(x+1) - 10x² + 4 = 0

2. Distribute the -3x² to the terms inside the second set of parentheses: 3x⁴ - 3x⁴ - 3x² - 10x² + 4 = 0

3. Combine like terms: -3x² - 10x² + 4 = 0

4. Combine the -3x² and -10x² terms: -13x² + 4 = 0

Now, you have a quadratic equation in the form ax² + bx + c = 0, where a = -13, b = 0, and c = 4. To solve for x, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, plug in the values of a, b, and c:

x = (0 ± √(0 - 4(-13)(4))) / (2(-13))

x = (± √(0 + 208)) / (-26)

x = (± √208) / (-26)

Now, simplify the expression further:

x = ± √(4 * 13) / (-26)

x = ± √(4) * √(13) / (-26)

x = ± 2√13 / (-26)

x = ± √13 / (-13)

So, the solutions for the equation 3x²(x-1)(x+1) - 10x² + 4 = 0 are:

x = -√13/13 and x = √13/13