Question

(x^2- x)^4 – 5((x^2 – x)^2)*x^2 + 6х^4 = 0​

Answer 1

Ответ:

Объяснение:

(x²-x)⁴-5(x²-x)²*x²+6x⁴ = 0

x⁴(x-1)⁴-5x²(x-1)²x²+ 6x⁴ = 0

x⁴((x-1)⁴-5(x-1)² + 6) = 0

x₁ = 0

(x-1)² = y

y²-5y+6 = 0

D = 25-24 = 1

y₁ = (5-1)/2 = 2

y₂ = (5+1)/2 = 3

(x-1)² = 2

x₂ = 1+$\sqrt{2}$

x₃ = 1-$\sqrt{2}$

(x-1)² = 3

x₄ = 1+$\sqrt{3}$

x₅ = 1-$\sqrt{3}$

Answer 2

To solve the given polynomial equation $(x^2 - x)^4 - 5((x^2 - x)^2)x^2 + 6x^4 = 0$, let's simplify the expression and then factorize it to find its roots.

Let's start by defining $y = x^2 - x$ to simplify the expression:

$y^4 - 5y^2x^2 + 6x^4 = 0$

Now, substitute $y^2$ with $z$ to simplify further:

$z^2 - 5xz + 6x^2 = 0$

Now, we can factorize this quadratic equation in $z$:

$z^2 - 5xz + 6x^2 = (z - 3x)(z - 2x) = 0$

Now, substitute $z = y^2$ back in:

$(y^2 - 3x)(y^2 - 2x) = 0$

Now, replace $y = x^2 - x$ again:

$((x^2 - x)^2 - 3x)((x^2 - x)^2 - 2x) = 0$

Now, we have two quadratic equations to solve for $x$:

1. $(x^2 - x)^2 - 3x = 0$
2. $(x^2 - x)^2 - 2x = 0$

Let's solve these two equations separately.

For the first equation:

$(x^2 - x)^2 - 3x = 0$

Expand and rearrange to obtain a quadratic equation in $x$:

$x^4 - 3x^3 + x^2 - 3x = 0$

We can factor this equation further:

$x(x^3 - 3x^2 + x - 3) = 0$

By using numerical methods or factoring, you can find the approximate values of $x$ that satisfy this equation.

For the second equation:

$(x^2 - x)^2 - 2x = 0$

Expand and rearrange to obtain a quadratic equation in $x$:

$x^4 - 2x^3 + x^2 - 2x = 0$

We can factor this equation further:

$x(x^3 - 2x^2 + x - 2) = 0$

Again, using numerical methods or factoring, you can find the approximate values of $x$ that satisfy this equation.

Please note that finding the exact solutions for these polynomial equations might be challenging, and numerical methods or computational tools might be needed to approximate the roots.