Question

(x^2-3x)^2+3(x^2 -3x)-28=0

Answer 1

Я запуталась, чуть начала мысль...

Answer 2

I can help you solve this equation. First, let's expand the brackets and simplify the expression:

(x^2-3x)^2+3(x^2-3x)-28=0
x^4-6x^3+9x^2+3x^2-9x-28=0
x^4-6x^3+12x^2-9x-28=0

Next, let's factorize the polynomial using the rational root theorem. We can try the possible factors of -28, which are ±1, ±2, ±4, ±7, ±14, and ±28. By plugging in these values, we can find that x=4 and x=-2 are roots of the equation. Therefore, we can divide the polynomial by (x-4) and (x+2) to get the other factors:

x^4-6x^3+12x^2-9x-28=0
(x-4)(x+2)(x^2-2x-7)=0

Finally, we can use the quadratic formula to find the remaining roots of the equation:

x^2-2x-7=0
x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-7)}}{2(1)}
x=\frac{2\pm\sqrt{36}}{2}
x=1\pm3
x=4 \text{ or } x=-2

Therefore, the four solutions of the equation are x=4, x=-2, x=4, and x=-2. We can write them as a set:

\{4,-2,4,-2\}

I hope this helps you understand how to solve this equation. If you have any other questions, feel free to ask me.Ответь+подробно.+(x^2-3x)^2+3(x^2+-3x)-28=0