Pregunta
upstudy study bank question image url

Solve for \( x \) and illustrate your answer on a number line. \( 5.1 \quad x^{2} \leq 64 \) \( 5.2 \quad 81-4 x^{2} \geq 0 \) \( 5.3 \quad x^{2}-x-12>0 \) \( 5.4 \quad x^{2}-7 x \geq 0 \)

Ask by Lee Hanson. in South Africa
Jan 23,2025

Solución de tutoría real

Respuesta verificada por el tutor

Responder

Here are the solutions to the inequalities: 1. **5.1. \( x^{2} \leq 64 \):** \( -8 \leq x \leq 8 \) 2. **5.2. \( 81 - 4x^{2} \geq 0 \):** \( -4.5 \leq x \leq 4.5 \) 3. **5.3. \( x^{2} - x - 12 > 0 \):** \( x < -3 \) or \( x > 4 \) 4. **5.4. \( x^{2} - 7x \geq 0 \):** \( x \leq 0 \) or \( x \geq 7 \) These solutions are illustrated on number lines showing the ranges of \( x \) that satisfy each inequality.

Solución

Sure! Let's solve each of the inequalities step by step and illustrate their solutions on a number line. --- ### **5.1. \( x^{2} \leq 64 \)** **Solution:** 1. **Solve the inequality:** \[ x^{2} \leq 64 \] Take the square root of both sides: \[ |x| \leq \sqrt{64} \\ |x| \leq 8 \] This implies: \[ -8 \leq x \leq 8 \] **Number Line Illustration:** ``` <----|========|========|========|========|----------> -8 0 8 ``` - The shaded region between -8 and 8 represents all real numbers \( x \) such that \( -8 \leq x \leq 8 \). - Solid circles at -8 and 8 indicate that these endpoints are included in the solution (since the inequality is ≤). --- ### **5.2. \( 81 - 4x^{2} \geq 0 \)** **Solution:** 1. **Rearrange the inequality:** \[ 81 - 4x^{2} \geq 0 \\ -4x^{2} \geq -81 \\ \] Multiply both sides by -1 (note that the inequality sign reverses when multiplying by a negative number): \[ 4x^{2} \leq 81 \\ x^{2} \leq \frac{81}{4} \\ x^{2} \leq 20.25 \] 2. **Take the square root of both sides:** \[ |x| \leq \sqrt{20.25} \\ |x| \leq 4.5 \] This implies: \[ -4.5 \leq x \leq 4.5 \] **Number Line Illustration:** ``` <----|=====|=====|=====|=====|-----> -4.5 0 4.5 ``` - The shaded region between -4.5 and 4.5 represents all real numbers \( x \) such that \( -4.5 \leq x \leq 4.5 \). - Solid circles at -4.5 and 4.5 indicate that these endpoints are included in the solution. --- ### **5.3. \( x^{2} - x - 12 > 0 \)** **Solution:** 1. **Factor the quadratic inequality:** \[ x^{2} - x - 12 > 0 \] Factor: \[ (x - 4)(x + 3) > 0 \] 2. **Find the critical points:** \[ x - 4 = 0 \Rightarrow x = 4 \\ x + 3 = 0 \Rightarrow x = -3 \] 3. **Determine the intervals to test:** The critical points divide the number line into three intervals: - \( x < -3 \) - \( -3 < x < 4 \) - \( x > 4 \) 4. **Test each interval:** - **Interval 1: \( x < -3 \)** (e.g., \( x = -4 \)) \[ (-4 - 4)(-4 + 3) = (-8)(-1) = 8 > 0 \quad \text{(True)} \] - **Interval 2: \( -3 < x < 4 \)** (e.g., \( x = 0 \)) \[ (0 - 4)(0 + 3) = (-4)(3) = -12 < 0 \quad \text{(False)} \] - **Interval 3: \( x > 4 \)** (e.g., \( x = 5 \)) \[ (5 - 4)(5 + 3) = (1)(8) = 8 > 0 \quad \text{(True)} \] 5. **Solution:** The inequality \( (x - 4)(x + 3) > 0 \) is satisfied in: \[ x < -3 \quad \text{or} \quad x > 4 \] **Number Line Illustration:** ``` <====| |==========| -3 4 ``` - Shaded regions extend to the left of -3 and to the right of 4. - Open circles at -3 and 4 indicate that these points are **not** included in the solution (since the inequality is >, not ≥). --- ### **5.4. \( x^{2} - 7x \geq 0 \)** **Solution:** 1. **Factor the quadratic inequality:** \[ x^{2} - 7x \geq 0 \\ x(x - 7) \geq 0 \] 2. **Find the critical points:** \[ x = 0 \\ x - 7 = 0 \Rightarrow x = 7 \] 3. **Determine the intervals to test:** The critical points divide the number line into three intervals: - \( x < 0 \) - \( 0 < x < 7 \) - \( x > 7 \) 4. **Test each interval:** - **Interval 1: \( x < 0 \)** (e.g., \( x = -1 \)) \[ (-1)(-1 - 7) = (-1)(-8) = 8 \geq 0 \quad \text{(True)} \] - **Interval 2: \( 0 < x < 7 \)** (e.g., \( x = 3 \)) \[ (3)(3 - 7) = (3)(-4) = -12 \geq 0 \quad \text{(False)} \] - **Interval 3: \( x > 7 \)** (e.g., \( x = 8 \)) \[ (8)(8 - 7) = (8)(1) = 8 \geq 0 \quad \text{(True)} \] 5. **Include critical points if they satisfy the inequality:** - At \( x = 0 \): \[ 0(0 - 7) = 0 \geq 0 \quad \text{(True)} \] - At \( x = 7 \): \[ 7(7 - 7) = 7(0) = 0 \geq 0 \quad \text{(True)} \] 6. **Solution:** The inequality \( x(x - 7) \geq 0 \) is satisfied in: \[ x \leq 0 \quad \text{or} \quad x \geq 7 \] **Number Line Illustration:** ``` <====|==========| 0 7 ``` - Shaded regions extend to the left of 0 and to the right of 7. - Solid circles at 0 and 7 indicate that these points are included in the solution (since the inequality is ≥). --- ### **Summary of Solutions:** 1. **5.1:** \( -8 \leq x \leq 8 \) 2. **5.2:** \( -4.5 \leq x \leq 4.5 \) 3. **5.3:** \( x < -3 \) or \( x > 4 \) 4. **5.4:** \( x \leq 0 \) or \( x \geq 7 \) These solutions are visually represented on their respective number lines, illustrating the ranges of \( x \) that satisfy each inequality.

Revisado y aprobado por el equipo de tutoría de UpStudy

error msg
Explicar
Simplifique esta solución

Mind Expander

Let’s tackle each inequality one at a time! ### 5.1: Solve \( x^{2} \leq 64 \) To solve for \( x \): \[ x^2 \leq 64 \implies -\sqrt{64} \leq x \leq \sqrt{64} \implies -8 \leq x \leq 8 \] **Number Line**: The solution is represented by a line segment from -8 to 8, including both endpoints. ### 5.2: Solve \( 81 - 4x^{2} \geq 0 \) Rearranging gives: \[ 4x^2 \leq 81 \implies x^2 \leq \frac{81}{4} \implies -\frac{9}{2} \leq x \leq \frac{9}{2} \] **Number Line**: The solution is shown with a line segment from -4.5 to 4.5, including both endpoints. ### 5.3: Solve \( x^{2} - x - 12 > 0 \) Factoring gives: \[ (x - 4)(x + 3) > 0 \] The critical points are \( x = 4 \) and \( x = -3 \). The solution intervals are \( (-\infty, -3) \cup (4, \infty) \). **Number Line**: The number line would show open intervals to the left of -3 and to the right of 4. ### 5.4: Solve \( x^{2} - 7x \geq 0 \) Factoring gives: \[ x(x - 7) \geq 0 \] The critical points are \( x = 0 \) and \( x = 7 \). Thus, the solution is \( x \in (-\infty, 0] \cup [7, \infty) \). **Number Line**: The solution is shown by shaded sections to the left of 0 and to the right of 7, including both 0 and 7. In summary, the solutions are: - 5.1: \([-8, 8]\) - 5.2: \([-4.5, 4.5]\) - 5.3: \( (-\infty, -3) \cup (4, \infty) \) - 5.4: \((- \infty, 0] \cup [7, \infty)\)

preguntas relacionadas

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad