Question

4x-x^4=0 2x^2-6x+4=0

Answer 1

Это система уравнения?

Answer 2

To solve the given equations $4x - x^4 = 0$ and $2x^2 - 6x + 4 = 0$, we can start by factoring the equations or by using other methods like the quadratic formula for the second equation and factoring for the first equation.

For the first equation $4x - x^4 = 0$:

Factor out $x$: $x(4 - x^3) = 0$

This equation has two solutions:

1. $x = 0$
2. $4 - x^3 = 0$

For the second equation $2x^2 - 6x + 4 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 2$, $b = -6$, and $c = 4$.

$x = \frac{6 \pm \sqrt{(-6)^2 - 4(2)(4)}}{2(2)}$ $x = \frac{6 \pm \sqrt{36 - 32}}{4}$ $x = \frac{6 \pm \sqrt{4}}{4}$

This gives us two solutions:

1. $x = \frac{6 + 2}{4} = 2$
2. $x = \frac{6 - 2}{4} = 1$

So, the solutions to the given equations are $x = 0$, $x = 2$, and $x = 1$.