To solve this inequality, we first need to find the critical points by setting the numerator and denominator equal to 0 and solving for x.
2x^2 + 3x - 2 = 02x−12x - 12x−1x+2x + 2x+2 = 0x = 1/2 or x = -2
2−x2-x2−x^29−x29-x^29−x2 = 02−x2-x2−x^2 = 0 or 9−x29-x^29−x2 = 0x = 2 or x = -2 or x = 3 or x = -3
Now, we need to create intervals using the critical points and test each interval to determine the sign of the expression.
Interval 1: −∞,−3-∞, -3−∞,−3 Choose x = -4:2−4-4−4^2 + 3−4-4−4 - 2 / (2−(−4))2(9−(−4)2)(2 -(-4))^2(9-(-4)^2)(2−(−4))2(9−(−4)2) = 50 / 144 > 0
Interval 2: −3,−2-3, -2−3,−2 Choose x = -2.5:2−2.5-2.5−2.5^2 + 3−2.5-2.5−2.5 - 2 / (2−(−2.5))2(9−(−2.5)2)(2-(-2.5))^2(9-(-2.5)^2)(2−(−2.5))2(9−(−2.5)2) = -0.8 / 36 > 0
Interval 3: −2,1/2-2, 1/2−2,1/2 Choose x = 0:2000^2 + 3000 - 2 / (2−0)2(9−02)(2-0)^2(9-0^2)(2−0)2(9−02) = -2 / 36 < 0
Interval 4: 1/2,21/2, 21/2,2 Choose x = 1:2111^2 + 3111 - 2 / (2−1)2(9−12)(2-1)^2(9-1^2)(2−1)2(9−12) = 3 / 7 > 0
Interval 5: 2,32, 32,3 Choose x = 2.5:22.52.52.5^2 + 32.52.52.5 - 2 / (2−2.5)2(9−2.52)(2-2.5)^2(9-2.5^2)(2−2.5)2(9−2.52) = 11.2 / 9.76 > 0
Interval 6: 3,∞3, ∞3,∞ Choose x = 4:2444^2 + 3444 - 2 / (2−4)2(9−42)(2-4)^2(9-4^2)(2−4)2(9−42) = 38 / 64 > 0
Therefore, the solution to the inequality 2x^2+3x-2/2−x2-x2−x^29−x29-x^29−x2 > 0 is x ∈ −∞,−3-∞, -3−∞,−3 U −2,1/2-2, 1/2−2,1/2 U 2,32, 32,3 U 3,∞3, ∞3,∞.
To solve this inequality, we first need to find the critical points by setting the numerator and denominator equal to 0 and solving for x.
2x^2 + 3x - 2 = 0
2x−12x - 12x−1x+2x + 2x+2 = 0
x = 1/2 or x = -2
2−x2-x2−x^29−x29-x^29−x2 = 0
2−x2-x2−x^2 = 0 or 9−x29-x^29−x2 = 0
x = 2 or x = -2 or x = 3 or x = -3
Now, we need to create intervals using the critical points and test each interval to determine the sign of the expression.
Interval 1: −∞,−3-∞, -3−∞,−3 Choose x = -4:
2−4-4−4^2 + 3−4-4−4 - 2 / (2−(−4))2(9−(−4)2)(2 -(-4))^2(9-(-4)^2)(2−(−4))2(9−(−4)2) = 50 / 144 > 0
Interval 2: −3,−2-3, -2−3,−2 Choose x = -2.5:
2−2.5-2.5−2.5^2 + 3−2.5-2.5−2.5 - 2 / (2−(−2.5))2(9−(−2.5)2)(2-(-2.5))^2(9-(-2.5)^2)(2−(−2.5))2(9−(−2.5)2) = -0.8 / 36 > 0
Interval 3: −2,1/2-2, 1/2−2,1/2 Choose x = 0:
2000^2 + 3000 - 2 / (2−0)2(9−02)(2-0)^2(9-0^2)(2−0)2(9−02) = -2 / 36 < 0
Interval 4: 1/2,21/2, 21/2,2 Choose x = 1:
2111^2 + 3111 - 2 / (2−1)2(9−12)(2-1)^2(9-1^2)(2−1)2(9−12) = 3 / 7 > 0
Interval 5: 2,32, 32,3 Choose x = 2.5:
22.52.52.5^2 + 32.52.52.5 - 2 / (2−2.5)2(9−2.52)(2-2.5)^2(9-2.5^2)(2−2.5)2(9−2.52) = 11.2 / 9.76 > 0
Interval 6: 3,∞3, ∞3,∞ Choose x = 4:
2444^2 + 3444 - 2 / (2−4)2(9−42)(2-4)^2(9-4^2)(2−4)2(9−42) = 38 / 64 > 0
Therefore, the solution to the inequality 2x^2+3x-2/2−x2-x2−x^29−x29-x^29−x2 > 0 is x ∈ −∞,−3-∞, -3−∞,−3 U −2,1/2-2, 1/2−2,1/2 U 2,32, 32,3 U 3,∞3, ∞3,∞.