Question

(X`2-3X-10)х(4 - 2X + 14/Х + 2)=0

Answer 1

Ответ:

Х≠-2

$\frac{(4 - 2x)(x + 2) + 14}{x + 2} = 0 \\ 4x + 8 - 2 {x}^{2} - 4x + 14 = 0 \\ 2 {x}^{2} = 22 \\ {x}^{2} = 11$

Х3, 4=±корень 11

Не поняла что должно быть в числитее, вот второй вариант

$\frac{4(x + 2) - 2x + 14}{x + 2} = 0 \\ 4x + 8 - 2x + 14 = 0 \\ 2x = - 22 \\ x3 = - 11$

${x}^{2} - 3x - 10 = 0 \\ \\ x1 + x2 = 3 \\ x1x2 = - 10 \\ \\ x1 = 5 \\ x2 = - 2$

Х2 не подходит

Answer 2

It looks like you're trying to solve a quadratic equation that is factored as a product of two expressions. The equation you've given is:

$(x^2 - 3x - 10) \cdot \left(4 - 2x + \frac{14}{x} + 2\right) = 0$

To solve this equation, you can set each factor equal to zero and solve for $x$:

1. Set $x^2 - 3x - 10 = 0$:

   This is a quadratic equation that can be factored or solved using the quadratic formula.

   Factoring: $(x - 5)(x + 2) = 0$ So, $x = 5$ or $x = -2$.

2. Set $4 - 2x + \frac{14}{x} + 2 = 0$:

   Combine like terms: $-2x + \frac{16}{x} + 6 = 0$

   Multiply both sides by $x$ to clear the fraction: $-2x^2 + 16 + 6x = 0$

   Rearrange: $-2x^2 + 6x + 16 = 0$

   This is another quadratic equation that can be factored or solved using the quadratic formula.

At this point, you have two sets of solutions:

1. From the first factor: $x = 5$ or $x = -2$.
2. From the second factor: Solve the quadratic equation $-2x^2 + 6x + 16 = 0$ to get two more solutions.

After solving the second quadratic equation, you should have a total of four potential solutions for $x$. Keep in mind that not all solutions might be valid for the original equation, as sometimes they could lead to denominators becoming zero (which would be undefined).