gmattesttaker2 wrote:Hello,

The following is from MGMAT Algebra Strategy Guide. (5th edition), p. 45. I am not getting the correct answer here. Can you please help?

x^3 < x^2 .Describe the possible values of x.

My approach:

x^3 < x^2
=> x^3 - x^2 < 0
=> x^2 (x - 1 ) < 0

Since x^2 > 0 => (x - 1) < 0 i.e. x < 1

x^2 > 0 => |x| > 0 => +x > 0 or -x > 0 i.e. x < 0


So now I have 3 values: x < 1 , x > 0 , x < 0

I think I am going wrong here. The book answer is [spoiler]"Any non-zero number less than 1"[/spoiler]


Can you please assist here? Thanks for your valuable time and help.

Best Regards,
Sri

Sri:

x^3 < x^2
=> x^3 - x^2 < 0
=> x^2 (x - 1 ) < 0
You are ok till this point.
First x^2 >=0. But we are given that product of (x^2) and (x-1) is negative. Hence x is neither equal to 0, nor equal to 1. Then (x^2) > 0. This is true regardless of any non-zero x. We do not need to consider this anymore.

So we are left with (x-1) < 0, which means x<1.

hence the answer is (-infinity< x < 0 and 0<x<1). In other words all values of x less than 1 (other than 0).

gmattesttaker2 wrote:Hello,

The following is from MGMAT Algebra Strategy Guide. (5th edition), p. 45. I am not getting the correct answer here. Can you please help?

x^3 < x^2 .Describe the possible values of x.

Can you please assist here? Thanks for your valuable time and help.

Best Regards,
Sri

Let's look at another approach we can take to the one posted by eagleeye.

First, a note of caution: when dealing with inequalities, be very wary of multiplying or dividing both sides by variables. Remember, if those variables turn out to be negative you have to reverse the inequality.

Second, if you know that you're dividing or multiplying by a positive value, it's safe to do so.

To quickly solve this particular question, we have to consider 2 cases:

1) x^2 is NOT = 0; and
2) x^2 IS = 0.

In the first case, we can simply divide both sides by x^2 (since anything squared is non-negative), giving us:

x < 1.

In the second case, we can see that x^2 = 0 doesn't fit this inequality (since 0 is NOT < 0).

Accordingly, x can be any value less than 1, EXCEPT 0. Done!

eagleeye wrote:

gmattesttaker2 wrote:Hello,

The following is from MGMAT Algebra Strategy Guide. (5th edition), p. 45. I am not getting the correct answer here. Can you please help?

x^3 < x^2 .Describe the possible values of x.

My approach:

x^3 < x^2
=> x^3 - x^2 < 0
=> x^2 (x - 1 ) < 0

Since x^2 > 0 => (x - 1) < 0 i.e. x < 1

x^2 > 0 => |x| > 0 => +x > 0 or -x > 0 i.e. x < 0


So now I have 3 values: x < 1 , x > 0 , x < 0

I think I am going wrong here. The book answer is [spoiler]"Any non-zero number less than 1"[/spoiler]


Can you please assist here? Thanks for your valuable time and help.

Best Regards,
Sri

Sri:

x^3 < x^2
=> x^3 - x^2 < 0
=> x^2 (x - 1 ) < 0
You are ok till this point.
First x^2 >=0. But we are given that product of (x^2) and (x-1) is negative. Hence x is neither equal to 0, nor equal to 1. Then (x^2) > 0. This is true regardless of any non-zero x. We do not need to consider this anymore.

So we are left with (x-1) < 0, which means x<1.

hence the answer is (-infinity< x < 0 and 0<x<1). In other words all values of x less than 1 (other than 0).

Hello Eagleeye,

Thank you very much for your excellent explanation (as always!). It is clear now.

Best Regards,
Sri

Stuart Kovinsky wrote:

gmattesttaker2 wrote:Hello,

The following is from MGMAT Algebra Strategy Guide. (5th edition), p. 45. I am not getting the correct answer here. Can you please help?

x^3 < x^2 .Describe the possible values of x.

Can you please assist here? Thanks for your valuable time and help.

Best Regards,
Sri

Let's look at another approach we can take to the one posted by eagleeye.

First, a note of caution: when dealing with inequalities, be very wary of multiplying or dividing both sides by variables. Remember, if those variables turn out to be negative you have to reverse the inequality.

Second, if you know that you're dividing or multiplying by a positive value, it's safe to do so.

To quickly solve this particular question, we have to consider 2 cases:

1) x^2 is NOT = 0; and
2) x^2 IS = 0.

In the first case, we can simply divide both sides by x^2 (since anything squared is non-negative), giving us:

x < 1.

In the second case, we can see that x^2 = 0 doesn't fit this inequality (since 0 is NOT < 0).

Accordingly, x can be any value less than 1, EXCEPT 0. Done!

Hello Stuart,

Thank you very much for your alternate approach. It is clear now. Thanks again.

Best Regards,
Sri