Question

1) x² - 7x-30<02) 4x²+16x>0​

Answer 1

Ответ:

Метод интервалов во 2 номере

Answer 2

Ответ:

1)x=(-3;10)

2)x=(-00;-4)U(0;+00)

$\infty$

Answer 3

To solve these inequalities, let's start with each one separately:

1. $x^2 - 7x - 30 < 0$

First, factor the quadratic:

$x^2 - 7x - 30 = (x - 10)(x + 3)$

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

$x - 10 = 0$ or $x + 3 = 0$

$x = 10$ or $x = -3$

These are the critical points. Now, create a number line and test the intervals between these critical points:

• Interval 1: $-\infty$ to $-3$

  • Choose a test point within this interval, for example, $x = -4$.
  • Plug it into the original inequality: $(-4 - 10)(-4 + 3) < 0$
  • Simplify: $(-14)(-1) < 0$
  • Since the product of two negative numbers is positive, this interval is part of the solution.

• Interval 2: $-3$ to $10$

  • Choose a test point within this interval, for example, $x = 0$.
  • Plug it into the original inequality: $(0 - 10)(0 + 3) < 0$
  • Simplify: $(-10)(3) < 0$
  • Since the product of a negative number and a positive number is negative, this interval is part of the solution.

• Interval 3: $10$ to $+\infty$

  • Choose a test point within this interval, for example, $x = 11$.
  • Plug it into the original inequality: $(11 - 10)(11 + 3) < 0$
  • Simplify: $(1)(14) < 0$
  • Since the product of a positive number and a positive number is positive, this interval is not part of the solution.

Now, combine the intervals where the inequality is true:

$x \in (-\infty, -3) \cup (-3, 10)$

So, the solution to $x^2 - 7x - 30 < 0$ is $x \in (-\infty, -3) \cup (-3, 10)$.

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2. $4x^2 + 16x > 0$

First, factor out the common factor of 4:

$4x^2 + 16x = 4x(x + 4)$

Now, consider the factors separately:

Factor 1: $4x$

Factor 2: $x + 4$

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

Factor 1: $4x = 0$ $x = 0$

Factor 2: $x + 4 = 0$ $x = -4$

These are the critical points. Now, create a number line and test the intervals between these critical points:

• Interval 1: $-\infty$ to $-4$

  • Choose a test point within this interval, for example, $x = -5$.
  • Plug it into the original inequality: $4(-5)(-5 + 4) > 0$
  • Simplify: $4(-5)(-1) > 0$
  • Since the product of two negative numbers is positive, this interval is part of the solution.

• Interval 2: $-4$ to $0$

  • Choose a test point within this interval, for example, $x = -2$.
  • Plug it into the original inequality: $4(-2)(-2 + 4) > 0$
  • Simplify: $4(-2)(2) > 0$
  • Since the product of a negative number and a positive number is negative, this interval is not part of the solution.

• Interval 3: $0$ to $+\infty$

  • Choose a test point within this interval, for example, $x = 1$.
  • Plug it into the original inequality: $4(1)(1 + 4) > 0$
  • Simplify: $4(1)(5) > 0$
  • Since the product of two positive numbers is positive, this interval is part of the solution.

Now, combine the intervals where the inequality is true:

$x \in (-\infty, -4) \cup (0, +\infty)$

So, the solution to $4x^2 + 16x > 0$ is $x \in (-\infty, -4) \cup (0, +\infty)$.