Question

(x^2+4x)(x^2+4x+17)+60=0​

Answer 1

Сначала раскрой скобки и вычисляю
А потом произведение равно нулю тогда и только тогда , когда один из множителей равен нулю , а другой при этом имеет смысл

Answer 2

To solve the quadratic equation $(x^2+4x)(x^2+4x+17)+60=0$, you can start by factoring and simplifying the equation:

$(x^2+4x)(x^2+4x+17)+60=0$

Now, let's factor the first part of the equation, $x^2+4x$, out:

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

Now, the equation becomes:

$(x(x+4))(x^2+4x+17)+60=0$

Next, let's simplify the second part of the equation, $x^2+4x+17$, by attempting to factor it. However, this quadratic does not have real roots because the discriminant ($b^2 - 4ac$) is negative, where $a = 1$, $b = 4$, and $c = 17$. Therefore, you cannot factor it further over the real numbers.

So, we have:

$(x(x+4))(x^2+4x+17)+60=0$

Now, you can't simplify the equation further by factoring. To solve for $x$, you may need to use numerical methods or a calculator. One approach is to use numerical methods like the quadratic formula or a graphing calculator to find approximate solutions.

Using the quadratic formula, you can find the roots:

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

In this case:

$a = 1$ $b = 4$ $c = 17$

Plugging in these values:

$$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(17)}}{2(1)}$$

$$x = \frac{-4 \pm \sqrt{16 - 68}}{2}$$

$$x = \frac{-4 \pm \sqrt{-52}}{2}$$

Since the discriminant is negative, there are no real solutions. The equation has complex solutions, which can be expressed as:

$$x = \frac{-4 \pm i\sqrt{52}}{2}$$

So, the solutions are complex numbers:

$$x = \frac{-4 + i\sqrt{52}}{2} \text{ and } x = \frac{-4 - i\sqrt{52}}{2}$$