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Assignment:Find the solution set of inequality 0.5 <= (3a^2+2a-sqrt(9a^4+6a^3+7a^2+4a+1)+1)/(3a^2) <= 2 .
Question type: Inequality
Solution:
The inequality can be reduced to 2 inequalities:
0.5 <= ( 3 * a ^ 2 + 2 * a - sqrt ( 9 * a ^ 4 + 6 * a ^ 3 + 7 * a ^ 2 + 4 * a + 1 ) + 1 ) / ( 3 * a ^ 2 ) (1)
( 3 * a ^ 2 + 2 * a - sqrt ( 9 * a ^ 4 + 6 * a ^ 3 + 7 * a ^ 2 + 4 * a + 1 ) + 1 ) / ( 3 * a ^ 2 ) <= 2 (2)
From the definition field of √
9 * x ^ 4 + 6 * x ^ 3 + 7 * x ^ 2 + 4 * x + 1 ≥ 0 (3 )
From the definition field of divisor
3 * x ^ 2 ≠ 0 (4 )
From inequality(1):
x ∈ Φ (Φ is empty)that is, the inequality will never be estatlished within the real number range!
From inequality(2):
a ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!
From inequality(3):
a ∈ R (R is all real numbers),that is, in the real number range, the inequality is always established!
From inequality(4):
a < 0 或 a > 0
From inequalities (1) and (2)
x ∈ Φ (Φ is empty)that is, the inequality will never be estatlished within the real number range! (5)
From inequalities (3) and (5)
x ∈ Φ (Φ is empty)that is, the inequality will never be estatlished within the real number range! (6)
From inequalities (4) and (6)
x ∈ Φ (Φ is empty)that is, the inequality will never be estatlished within the real number range! (7)
The final solution set is :
x ∈ Φ (Φ is empty)that is, the inequality will never be estatlished within the real number range!Your problem has not been solved here? Please take a look at the hot problems !