Question

(x^2+3x+1)(x^2+3x+3)-3=0

Answer 1

Ответ:

{ -3 ; 0 }

Объяснение:

(x^2+3x+1)(x^2+3x+3)-3=0

Пусть x^2+3x+1  = t

t(t+2)-3=0

t²+2t-3=0

по теореме виета:

t1 = 1

t2 = -3

обратная замена:

x^2+3x+1=1

x^2+3x=0

x(x+3)=0

x=0

x= -3

x^2+3x+1= -3

x^2+3x+4=0

D = 9-16 = -7

D<0 , нет действительных корней

Answer 2

To solve the quadratic equation $(x^2+3x+1)(x^2+3x+3)-3=0$, let's simplify the expression first and then solve for $x$.

Let $y = x^2 + 3x$. Then the equation becomes:

$(y + 1)(y + 3) - 3 = 0$

Expanding this expression:

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

Combine like terms:

$y^2 + 4y = 0$

Factor out the common term:

$y(y + 4) = 0$

Now, substitute back $y = x^2 + 3x$:

$(x^2 + 3x)(x^2 + 3x + 4) = 0$

Now, set each factor equal to zero and solve for $x$:

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

2. $x^2 + 3x + 4 = 0$ (Use the quadratic formula since this doesn't factor easily.)

   The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where the quadratic equation is $ax^2 + bx + c = 0$.

   For $x^2 + 3x + 4 = 0$, $a = 1$, $b = 3$, and $c = 4$.

   $x = \frac{-3 \pm \sqrt{3^2 - 4(1)(4)}}{2(1)}$ $x = \frac{-3 \pm \sqrt{-7}}{2}$

   Since the discriminant ($b^2 - 4ac$) is negative, the solutions are complex numbers. Therefore, the roots are complex conjugates:

   $x = \frac{-3}{2} \pm \frac{\sqrt{7}}{2}i$

So, the solutions to the equation $(x^2+3x+1)(x^2+3x+3)-3=0$ are $x = 0$, $x = -3$, $x = \frac{-3}{2} + \frac{\sqrt{7}}{2}i$, and $x = \frac{-3}{2} - \frac{\sqrt{7}}{2}i$.