Question

(x^2+x)^2 + (x^2 + x)-2 = 0.​

Answer 1

Ответ:

$x_1=\frac{-1+\sqrt{5} }{2} ; x_2=\frac{-1-\sqrt{5} }{2}$

Пошаговое объяснение:

(x²+x)²+ (x² + x)-2 = 0.

x² + x=t

t²+t-2=0

$t=\frac{-1+-\sqrt{1+8} }{2}=\frac{-1+-\sqrt{9} }{2} =\frac{-1+-3}{2} \\\\ t_1=\frac{2}{2}=1;\\ t_2=\frac{-4}{2} =-2\\\\x^2+x=1\\\\x^2+x-1=0\\\\x=\frac{-1+-\sqrt{1+4} }{2} =\frac{-1+-\sqrt{5} }{2} ;\\\\x_1=\frac{-1+\sqrt{5} }{2} ; x_2=\frac{-1-\sqrt{5} }{2} \\\\x^2+x=-2\\\\x^2+x+2=0\\\\D=1-8$

Answer 2

$(x^{2}+x)^{2} + (x^{2}+x)-2=0$

Замена: $x^{2}+x=t$

$t^{2}+t-2=0$

$t_{1}=-2; \ t_{2} = 1$

Обратная замена:

$1) \ x^{2} + x =-2\\x^{2}+x+2=0\\D = 1^{2} - 4 \cdot 1 \cdot 2$

Нет корней

$2) \ x^{2}+x=1\\x^{2}+x-1=0\\D=1^{2}-4\cdot 1 \cdot(-1) =5$

$x_{1,2} = \dfrac{-1\pm \sqrt{5}}{2}$

Ответ: $x_{1,2} = \dfrac{-1\pm \sqrt{5}}{2}$

Answer 3

To solve the quadratic equation $(x^2+x)^2 + (x^2 + x) - 2 = 0$, we can follow these steps:

1. Distribute the terms inside the parentheses: $(x^4 + 2x^3 + x^2) + (x^2 + x) - 2 = 0$.

2. Combine like terms: $x^4 + 2x^3 + 2x^2 + x - 2 = 0$.

3. Rearrange the equation in descending order of powers of $x$: $x^4 + 2x^3 + 2x^2 + x - 2 = 0$.

4. To solve this equation, you can use numerical methods or graphing software, as it doesn't have a simple algebraic solution.

If you're looking for numerical solutions, you can use methods like the Newton-Raphson method or graphing tools to find the approximate solutions.