The solution to the inequality is D. x∈(−∞,−4)∪( 5/2 ,∞). The correct answer is D. x∈(−∞,−4)∪(5/2 ,∞).
To solve the quadratic inequality (x+4)(2x−5)>0, let's find the critical points by setting each factor equal to zero:
x+4=0 gives x=−4.
2x−5=0 gives x= 5/2.
These critical points divide the number line into three intervals:
(−∞,−4),
(−4, 5/2 ), and ( 5/2 ,+∞).
Now, let's test a value in each interval to determine where the inequality holds true:
Test a value less than −4, say x=−5.
(−5+4)(2(−5)−5)=(−1)(−15)=15>0. So, the inequality holds in (−∞,−4).
Test a value between −4 and 5/2 , say x=0.
(0+4)(2(0)−5)=(4)(−5)=−20<0. So, the inequality doesn't hold in (−4, 5/2 ).
Test a value greater than 5/2 , say x=3.
(3+4)(2(3)−5)=(7)(1)=7>0. So, the inequality holds in (5/2 ,+∞).
Therefore, the solution to the inequality is D. x∈(−∞,−4)∪( 5/2 ,∞).
The correct answer is D. x∈(−∞,−4)∪(5/2 ,∞)
Question
Solve the following quadratic inequality.
(x+4) (2x - 5) > 0
A. IE (-∞, -4) U (7, ∞0)
B. x€ (-∞, -4) U (1, 00)
C. SE (-∞, -6) U (2/3, 0.)
D. x€ (-∞, -4) U (5/2,∞)