Ask Question
16 June, 04:31

9 • 8 - 2x + x2 ≥ x3 + x2

+5
Answers (2)
  1. 16 June, 04:37
    0
    9 (8) - 2x+x^2>=x^3+x^2

    -2x+72-x^3>=0

    -x^3-2x+72>=0

    X^3+2x-72<=0

    Factor

    (X-4) (x^2+4x+18) <=0

    Solve for x

    X<=4
  2. 16 June, 04:58
    0
    x² - 2x + 9 (8) ≥ x³ + x²

    x² - 2x + 72 ≥ x³ + x²

    - x² - x²

    -x³ - 2x + 72 ≥ 0

    -1 (x³) - 1 (2x) - 1 (72) ≥ 0

    -1 (x³ + 2x - 72) ≥ 0

    -1 - 1

    x³ + 2x - 72 ≥ 0

    x³ + 4x² - 4x² + 18x - 16x - 72 ≥ 0

    x³ + 4x² + 18x - 4x² - 16x - 72 ≥ 0

    x (x²) + x (4x) + x (18) - 4 (x²) - 4 (4x) - 4 (18) ≥ 0

    x (x² + 4x + 18) - 4 (x² + 4x + 18) ≥ 0

    (x - 4) (x² + 4x + 18) ≥ 0

    x - 4 ≥ 0 or x² + 4x + 18 ≥ 0

    + 4 + 4 x ≥ - (4) ± √ ((4) ² - 4 (1) (18))

    x ≥ 4 2 (1)

    x ≥ - 4 ± √ (16 - 72)

    2

    x ≥ - 4 ± √ (-56)

    2

    x ≥ - 4 ± 2i√ (14)

    2

    x ≥ - 2 ± i√ (14)

    x ≥ - 2 + i√ (14) or x ≥ - 2 - i√ (14)

    Solution Set: {4, - 2 ± i√ (14) }
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “9 • 8 - 2x + x2 ≥ x3 + x2 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers