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10 Jul 2017, 11:51
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If you look at x^3 < X^2<0, then it has solution X<1 except X=0
but when same equation is rearranged to:
X(X^2 -1)<0, the solution is 0<X<1 and X< -1. For -1<X<0, the equation is not satisfied as it yields positive value, eg x= -1/2, then X(X^2 -1)<0 yeilds
positive value, therefore solution -1<X<0 is not possible. But from equation x^3 < X^2, which is the same the equation, the solution includes the value -1<X<0
as well.
So the equation is same and when you arrange it the solutions are different why?...(is it due to some sign constraints)
but when same equation is rearranged to:
X(X^2 -1)<0, the solution is 0<X<1 and X< -1. For -1<X<0, the equation is not satisfied as it yields positive value, eg x= -1/2, then X(X^2 -1)<0 yeilds
positive value, therefore solution -1<X<0 is not possible. But from equation x^3 < X^2, which is the same the equation, the solution includes the value -1<X<0
as well.
So the equation is same and when you arrange it the solutions are different why?...(is it due to some sign constraints)
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10 Jul 2017, 16:06
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if you start from this inequality:
x^3 < x^2
and subtract x^2 on both sides, you get
x^3 - x^2 < 0
Now you can factor out x^2 to get:
x^2 ( x - 1) < 0
You'll find this inequality has the same solutions as the original one. There's an error in the factorization in your post, which is the reason your two inequalities have different solutions - you arrived at x (x^2 - 1) < 0, which is not the same as your original inequality.
x^3 < x^2
and subtract x^2 on both sides, you get
x^3 - x^2 < 0
Now you can factor out x^2 to get:
x^2 ( x - 1) < 0
You'll find this inequality has the same solutions as the original one. There's an error in the factorization in your post, which is the reason your two inequalities have different solutions - you arrived at x (x^2 - 1) < 0, which is not the same as your original inequality.
10 Jul 2017, 16:09
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Incidentally, there's no advantage in this particular question to doing any rearrangement. If we know:
x^3 < x^2
we can just divide both sides by x^2; because x^2 can't be negative, we don't need to worry about whether to flip the inequality. So we get:
x < 1
But since we're not allowed to divide by 0, we have to check whether x = 0 ought to be a solution too, by plugging it into the original inequality, and it shouldn't be, so the solutions are 0 < x < 1 and x < 0.
x^3 < x^2
we can just divide both sides by x^2; because x^2 can't be negative, we don't need to worry about whether to flip the inequality. So we get:
x < 1
But since we're not allowed to divide by 0, we have to check whether x = 0 ought to be a solution too, by plugging it into the original inequality, and it shouldn't be, so the solutions are 0 < x < 1 and x < 0.
