Find the set of values for which 6 + x ≥ x
2
and x + 2 < x
2
How do I do this A level Maths q?
2020-06-22 12:29 am
回答 (2)
2020-06-22 12:57 am
6 + x ≥ x² and x + 2 < x²
x² ≤ 6 + x and x² > x + 2
x² - x - 6 ≤ 6 + x - x - 6 and x² - x - 2 > x + 2 - x - 2
x² - x - 6 ≤ 0 and x² - x - 2 > 0
(x + 2)(x - 3) ≤ 0 and (x + 1)(x - 2) > 0
-2 ≤ x ≤ 3 and (x < -1 or x > 2)
Refer to the number line below.
Range of x: -2 ≤ x < -1 or 2 < x ≤ 3
x² ≤ 6 + x and x² > x + 2
x² - x - 6 ≤ 6 + x - x - 6 and x² - x - 2 > x + 2 - x - 2
x² - x - 6 ≤ 0 and x² - x - 2 > 0
(x + 2)(x - 3) ≤ 0 and (x + 1)(x - 2) > 0
-2 ≤ x ≤ 3 and (x < -1 or x > 2)
Refer to the number line below.
Range of x: -2 ≤ x < -1 or 2 < x ≤ 3
2020-06-22 12:34 am
Assuming you mean 6 + x ≥ x^2 and x + 2 < x^2,
6 + x ≥ x^2
0 ≥ x^2 - x - 6
0 ≥ (x - 3)(x + 2)
-2 ≤ x ≤ 3
x ∈ [-2, 3]
x + 2 < x^2
0 < x^2 - x - 2
0 < (x - 2)(x + 1)
x > 2 or x < -1
x ∈ (-∞ , -1) U (2,∞ )
x ∈ [-2, 3] ∩ ( (-∞ , -1) U (2,∞ ) )
x ∈ [-2, -1) U (2, 3]
6 + x ≥ x^2
0 ≥ x^2 - x - 6
0 ≥ (x - 3)(x + 2)
-2 ≤ x ≤ 3
x ∈ [-2, 3]
x + 2 < x^2
0 < x^2 - x - 2
0 < (x - 2)(x + 1)
x > 2 or x < -1
x ∈ (-∞ , -1) U (2,∞ )
x ∈ [-2, 3] ∩ ( (-∞ , -1) U (2,∞ ) )
x ∈ [-2, -1) U (2, 3]
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