Question

При каких значениях х верно неравенство x2(икс в квадрате) - 2x − 3 > 0

Answer 1

$x^2-2x-3>0$

$(x-3)(x+1)>0$

$x\in (-\infty;-1)u(3;\infty)$

Answer 2

I can help you with math problems and other topics.

To solve this inequality, we need to find the values of $x$ that make the expression $$x^2 - 2x - 3$$ positive. One way to do this is to factor the expression as $$(x - 3)(x + 1)$$ and then use the sign rule for products. The sign rule says that a product of two factors is positive if and only if both factors have the same sign. Therefore, we have two cases:

- Case 1: Both factors are positive, i.e. $x - 3 > 0$ and $x + 1 > 0$. This implies that $x > 3$ and $x > -1$, which can be combined as $x > 3$. - Case 2: Both factors are negative, i.e. $x - 3 < 0$ and $x + 1 < 0$. This implies that $x < 3$ and $x < -1$, which can be combined as $x < -1$.

Hence, the solution set for the inequality is $$x < -1 \text{ or } x > 3$$

You can check your answer by plugging in some values of $x$ into the original expression and see if they satisfy the inequality. For example, if $x = -2$, then $$x^2 - 2x - 3 = (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5 > 0$$ which is true. If $x = 2$, then $$x^2 - 2x - 3 = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3 < 0$$ which is false.Ответь+подробно.+При+каких+значениях+х+верно+неравенство+x2(икс+в+квадрате)+-+2x+−+3+>+0