INECUATII CU MODUL GRADUL I IN MULTIMEA NUMERELOR REALE CLASA A 8 A MATEMATICA EXERCITII REZOLVATE

Invata Matematica Usor2 minutes read

The text explains how to solve equations and inequalities involving the modulus function by isolating the expression and considering different cases based on the sign of the expression within the modulus. It emphasizes understanding the definitions of closed and open intervals and the manipulation of inequalities to find the appropriate solutions.

Insights

  • The concept of modulo is essential in solving absolute value equations and inequalities, as demonstrated by the equations \( |2x 3| = 7 \) and \( |2x - 3| \leq 7 \), where different cases lead to distinct solutions; for instance, \( |2x - 3| = 7 \) results in \( x \in \{2, 5\} \) while the inequality \( |2x - 3| \leq 7 \) leads to a continuous range of solutions \( x \in [-2, 5] \).
  • A systematic approach to solving inequalities with absolute values involves breaking them into cases based on the inequality sign, as seen in the cases for \( |x + 1| \geq 3 \) and \( |2x 5| > 5 \); this method allows for clear identification of solution sets, such as \( x \in (-\infty, -4] \cup [2, \infty) \) for the former and \( x \in (-\infty, 0) \cup (5, \infty) \) for the latter, highlighting the importance of understanding both closed and open intervals in the context of inequalities.

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Recent questions

  • What is the definition of a modulus?

    The modulus of a real number is a mathematical function that measures its distance from zero on the number line. Specifically, for any real number \( x \), the modulus is defined as \( x \) if \( x \) is greater than or equal to zero, and as \( -x \) if \( x \) is less than zero. This means that the modulus always yields a non-negative result, effectively representing the absolute value of the number. Understanding this concept is fundamental in various mathematical applications, including solving equations and inequalities that involve absolute values.

  • How do I solve absolute value equations?

    To solve absolute value equations, you typically start by isolating the absolute value expression on one side of the equation. Once isolated, you can set up two separate equations to account for the two possible cases: one where the expression inside the absolute value is equal to the positive value and another where it is equal to the negative value. For example, if you have an equation like \( |2x - 3| = 7 \), you would create two equations: \( 2x - 3 = 7 \) and \( 2x - 3 = -7 \). Solving these equations will yield the possible solutions for \( x \), which you can then verify by substituting back into the original equation.

  • What are the steps to solve inequalities with moduli?

    Solving inequalities that involve moduli requires a systematic approach. First, you need to rewrite the inequality in a form that isolates the absolute value expression. Depending on the inequality sign, you will then split the problem into different cases. For instance, if you have an inequality like \( |x + 1| \geq 3 \), you would break it down into two separate inequalities: \( x + 1 \leq -3 \) and \( x + 1 \geq 3 \). After solving these inequalities, you combine the results to find the solution set, which may include intervals or specific values. It’s important to pay attention to whether the inequality is strict (greater than or less than) or inclusive (greater than or equal to, less than or equal to) as this affects the endpoints of your solution.

  • What does it mean for an inequality to be strict?

    A strict inequality refers to an inequality that does not include the endpoints of the range it describes. In mathematical terms, this means using symbols like \( < \) or \( > \) instead of \( \leq \) or \( \geq \). For example, if you have an inequality such as \( x > 5 \), it indicates that \( x \) can take any value greater than 5, but not equal to 5 itself. This distinction is crucial when graphing the solution on a number line, as strict inequalities are represented with open circles at the endpoints, while inclusive inequalities are represented with closed circles. Understanding the difference between strict and inclusive inequalities is essential for accurately interpreting and solving mathematical problems.

  • How do I find the solution set for inequalities?

    To find the solution set for inequalities, you first need to manipulate the inequality to isolate the variable. This often involves adding, subtracting, multiplying, or dividing both sides of the inequality by the same number, while being cautious about the direction of the inequality sign, especially when multiplying or dividing by a negative number. Once the variable is isolated, you can express the solution in interval notation or as a set of values. For example, if you solve an inequality and find that \( -2 \leq x \leq 5 \), the solution set can be expressed as the interval \( [-2, 5] \). It’s important to check your solutions by substituting values back into the original inequality to ensure they satisfy the condition.

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Summary

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Solving Inequalities with Modulo Functions

  • The modulo of a real number \( x \) is defined as \( x \) if \( x \geq 0 \) and \( -x \) if \( x < 0 \). This concept is applied to solve the equation \( |2x - 3| = 7 \) by considering two cases: \( 2x - 3 = 7 \) and \( 2x - 3 = -7 \). The first case yields \( x = 5 \) and the second case yields \( x = 2 \), resulting in the solutions \( x \in \{2, 5\} \).
  • For the inequality \( |2x - 3| \leq 7 \), it can be rewritten as \( -7 \leq 2x - 3 \leq 7 \). Adding 3 to all parts gives \( -4 \leq 2x \leq 10 \). Dividing by 2 results in \( -2 \leq x \leq 5 \), leading to the solution \( x \in [-2, 5] \).
  • In the case of the inequality \( \frac{|2x + 1|}{9} \leq 1 \), multiplying through by 9 gives \( -9 \leq 2x + 1 \leq 9 \). Subtracting 1 results in \( -10 \leq 2x \leq 8 \). Dividing by 2 yields \( -5 \leq x \leq 4 \), thus the solution is \( x \in [-5, 4] \).
  • For the inequality \( |x + 1| \geq 3 \), it can be split into two cases: \( x + 1 \leq -3 \) and \( x + 1 \geq 3 \). Solving these gives \( x \leq -4 \) and \( x \geq 2 \), resulting in the solution \( x \in (-\infty, -4] \cup [2, \infty) \).
  • The inequality \( |2x - 5| > 5 \) leads to two cases: \( 2x - 5 < -5 \) and \( 2x - 5 > 5 \). The first case simplifies to \( 2x < 0 \) or \( x < 0 \), while the second case simplifies to \( 2x > 10 \) or \( x > 5 \). Therefore, the solution is \( x \in (-\infty, 0) \cup (5, \infty) \).
  • The general approach for solving inequalities involving moduli includes isolating the expression within the modulus, determining the appropriate intervals based on the inequality sign, and solving for \( x \) accordingly.
  • To effectively solve any inequality involving moduli, it is crucial to understand the definitions of closed and open intervals, as well as how to manipulate inequalities through addition, subtraction, multiplication, and division while maintaining the integrity of the inequality signs.
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