Question

(x^2+x+1)(x^2+x+2)-6=0

Answer 1

Ответ:

решение на фотографии

Answer 2

$(x^{2}+x+1)(x^{2}+x+2)-6=0\\\\x^{2}+x+1=m, \ \Rightarrow \ x^{2}+x+2=m+1\\\\m*(m+1)-6=0\\\\m^{2}+m-6=0\\\\m_{1}=-3\\\\m_{2}=2- \ teorema \ Vieta\\\\1)x^{2}+x+1=-3\\\\x^{2}+x+4=0\\\\D=1^{2}-4*4=1-16=-15$

Answer 3

To solve the equation $(x^2+x+1)(x^2+x+2) - 6 = 0$, you can follow these steps:

1. Expand the expressions:

   $(x^2+x+1)(x^2+x+2) = x^4 + 2x^3 + 3x^2 + 3x + 2$

2. Rewrite the equation with the expanded expressions:

   $x^4 + 2x^3 + 3x^2 + 3x + 2 - 6 = 0$

3. Combine like terms:

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

4. Try to factor the equation or find its roots. In this case, factoring might not be straightforward, so you can use numerical methods or a calculator to find the roots.

Using numerical methods or a graphing calculator, you can find the approximate solutions to this equation:

• $x \approx -1.8089$
• $x \approx -0.5774$
• $x \approx 1.6932$
• $x \approx 1.6932$ (This solution repeats due to a double root)

These are approximate solutions rounded to four decimal places. Keep in mind that the actual decimal representation of these solutions may have more decimal places if you need greater precision.