Math Problem Statement

Solve the following quadratic inequalities presented in the image:

Solution

The image shows a set of quadratic inequalities to be solved. The first few inequalities are solved as examples in the table, and there are additional exercises below. Let me guide you through solving these inequalities one by one:

  1. x20x^2 \leq 0
    The inequality x20x^2 \leq 0 is satisfied only when x=0x = 0, since x2x^2 is always non-negative and equal to 0 only at x=0x = 0.
    Solution: x=0x = 0.

  2. x20-x^2 \geq 0
    This is equivalent to x20x^2 \leq 0, which, like the previous one, is only true when x=0x = 0.
    Solution: x=0x = 0.

  3. x2+3x180x^2 + 3x - 18 \leq 0
    First, solve the quadratic equation x2+3x18=0x^2 + 3x - 18 = 0 using the quadratic formula:
    x=3±324(1)(18)2(1)=3±9+722=3±812=3±92x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-18)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 72}}{2} = \frac{-3 \pm \sqrt{81}}{2} = \frac{-3 \pm 9}{2} So, x1=3x_1 = 3 and x2=6x_2 = -6. Now, we analyze the intervals:

    • For x<6x < -6, x2+3x18>0x^2 + 3x - 18 > 0.
    • For 6<x<3-6 < x < 3, x2+3x180x^2 + 3x - 18 \leq 0.
    • For x>3x > 3, x2+3x18>0x^2 + 3x - 18 > 0.
      Solution: 6x3-6 \leq x \leq 3.
  4. 2x2+7x<02x^2 + 7x < 0
    Solve the quadratic equation 2x2+7x=02x^2 + 7x = 0 by factoring:
    x(2x+7)=0x(2x + 7) = 0 So, x=0x = 0 and x=72x = -\frac{7}{2}. Now, analyze the intervals:

    • For x<72x < -\frac{7}{2}, 2x2+7x>02x^2 + 7x > 0.
    • For 72<x<0-\frac{7}{2} < x < 0, 2x2+7x<02x^2 + 7x < 0.
    • For x>0x > 0, 2x2+7x>02x^2 + 7x > 0.
      Solution: 72<x<0-\frac{7}{2} < x < 0.
  5. x2+160x^2 + 16 \leq 0
    Since x2x^2 is always non-negative and x2+16x^2 + 16 is strictly greater than 0 for all real values of xx, there is no real solution to this inequality.
    Solution: No solution.

  6. 3x25x20-3x^2 - 5x - 2 \leq 0
    Solve the quadratic equation 3x25x2=0-3x^2 - 5x - 2 = 0 using the quadratic formula:
    x=(5)±(5)24(3)(2)2(3)=5±25246=5±16x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-3)(-2)}}{2(-3)} = \frac{5 \pm \sqrt{25 - 24}}{-6} = \frac{5 \pm 1}{-6} So, x1=1x_1 = -1 and x2=23x_2 = -\frac{2}{3}. Now, analyze the intervals:

    • For x<1x < -1, 3x25x20-3x^2 - 5x - 2 \leq 0.
    • For 1x23-1 \leq x \leq -\frac{2}{3}, 3x25x20-3x^2 - 5x - 2 \leq 0.
    • For x>23x > -\frac{2}{3}, 3x25x2>0-3x^2 - 5x - 2 > 0.
      Solution: 1x23-1 \leq x \leq -\frac{2}{3}.
  7. 3x2+1203x^2 + 12 \leq 0
    Like in problem 5, since 3x2+123x^2 + 12 is always positive, there is no real solution.
    Solution: No solution.

Would you like any specific details or clarifications for these solutions? Here are some related questions:

  1. How do you find the solutions for quadratic inequalities with two real roots?
  2. Can you explain the graphical interpretation of quadratic inequalities?
  3. What is the difference between strict and non-strict inequalities in quadratic equations?
  4. Why do some quadratic inequalities have no solution?
  5. How can you solve inequalities using the quadratic formula?

Tip: Always plot the quadratic function to better understand the regions where the inequality holds.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Inequalities
Factoring
Quadratic Formula

Formulas

Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
Factoring quadratics: ax² + bx + c = 0

Theorems

Quadratic formula
Sign of quadratic expressions
Interval method

Suitable Grade Level

Grades 9-12