The questions is simply: Find the range of values of x for which x(2-x)<-15
This is what i did:
x(2-x)+15<0
2x - x^2 + 15 < 0
-x^2 + 2x + 15 < 0
(x+3)(-x+15) < 0
-3<x<5
but when i check by subbing the number, say, 4 into x, the answer is always greater than zero. And the correct answer should be x<3 or x>5, which also seems wrong, because when i sub a number, say 2, into the inequality, the answer is still greater than 0.
what is wrong??? I've checked the things i did so many times and i don't see anything wrong????
Look at this:
(x+3)<0
x<-3
Not greater than.
-x+5<0
x>5
so x<-3 and x>5
The mistake in your calculations lies in the step where you factor the quadratic expression. The correct factorization is actually:
x + 3 > 0 and -x + 5 > 0
This is because you want the product of the two factors to be negative, and for that to be true, one factor should be negative and the other positive. By setting each factor greater than zero, you ensure that they have opposite signs.
Now, let's solve each inequality separately:
For x + 3 > 0, subtracting 3 from both sides gives x > -3.
For -x + 5 > 0, adding x to both sides and subtracting 5 from both sides gives x < 5.
When you combine both inequality solutions, you get -3 < x < 5. This is the correct range of values for x when x(2-x) < -15.
To check whether this is correct, you can substitute some numbers within this range into the inequality:
If you substitute x = 0, you get 0(2-0) < -15, which simplifies to 0 < -15, which is false.
If you substitute x = -4, you get -4(2-(-4)) < -15, which simplifies to -4(6) < -15, which is true (-24 < -15).
Therefore, the correct solution is x < -3 or x > 5.