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If x is positive, is x > 3 [#permalink]
16 Apr 2012, 08:42

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Question Stats:

50% (01:45) correct
50% (00:53) wrong based on 129 sessions

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9

Can someone point a mistake in my method? (1) Taking one of the equations: (x - 1)^2 > 4 x^2 + 1 - 2x > 4 x^2 + 1 - 2x - 4 > 0 x^2 - 3x + 1x - 3 > 0 (x-3) (x+1) > 0 x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Re: If x is positive, is x > 3 [#permalink]
16 Apr 2012, 10:58

3

This post received KUDOS

Expert's post

rohitgoel15 wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9

Can someone point a mistake in my method? (1) Taking one of the equations: (x - 1)^2 > 4 x^2 + 1 - 2x > 4 x^2 + 1 - 2x - 4 > 0 x^2 - 3x + 1x - 3 > 0 (x-3) (x+1) > 0 x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> (x+1)(x-3)>0 --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> (x+1)(x-5)>0 --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: |x-1|>2. |x-1| is just the distance between 1 and x on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> |x-2|>3. The same here: |x-2| is just the distance between 2 and x on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

Re: If x is positive, is x > 3 [#permalink]
20 Jul 2013, 05:57

Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?

Re: If x is positive, is x > 3 [#permalink]
20 Jul 2013, 06:32

Expert's post

fozzzy wrote:

Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?

If it were (x - 1)^2 > -4, it would simply mean that x can take any value.

As for general rules for inequalities: taking the square root, squaring, ...

ADDING/SUBTRACTING INEQUALITIES:

You can only add inequalities when their signs are in the same direction:

If a>b and c>d (signs in same direction: > and >) --> a+c>b+d. Example: 3<4 and 2<5 --> 3+2<4+5.

You can only apply subtraction when their signs are in the opposite directions:

If a>b and c<d (signs in opposite direction: > and <) --> a-c>b-d (take the sign of the inequality you subtract from). Example: 3<4 and 5>1 --> 3-5<4-1.

RAISING INEQUALITIES TO EVEN/ODD POWER:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: 2<4 --> we can square both sides and write: 2^2<4^2; 0\leq{x}<{y} --> we can square both sides and write: x^2<y^2;

But if either of side is negative then raising to even power doesn't always work. For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3; x<y --> we can raise both sides to third power and write: x^3<y^3.

Re: If x is positive, is x > 3 [#permalink]
05 Sep 2013, 02:47

Bunuel wrote:

rohitgoel15 wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9

Can someone point a mistake in my method? (1) Taking one of the equations: (x - 1)^2 > 4 x^2 + 1 - 2x > 4 x^2 + 1 - 2x - 4 > 0 x^2 - 3x + 1x - 3 > 0 (x-3) (x+1) > 0 x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> (x+1)(x-3)>0 --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> (x+1)(x-5)>0 --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: |x-1|>2. |x-1| is just the distance between 1 and x on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> |x-2|>3. The same here: |x-2| is just the distance between 2 and x on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

Answer: D.

I used x= +6 and -6 ..Which is true in both the cases..it shud be E..

Re: If x is positive, is x > 3 [#permalink]
05 Sep 2013, 02:52

Expert's post

SUNGMAT710 wrote:

Bunuel wrote:

rohitgoel15 wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9

Can someone point a mistake in my method? (1) Taking one of the equations: (x - 1)^2 > 4 x^2 + 1 - 2x > 4 x^2 + 1 - 2x - 4 > 0 x^2 - 3x + 1x - 3 > 0 (x-3) (x+1) > 0 x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> (x+1)(x-3)>0 --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> (x+1)(x-5)>0 --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: |x-1|>2. |x-1| is just the distance between 1 and x on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> |x-2|>3. The same here: |x-2| is just the distance between 2 and x on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

Answer: D.

I used x= +6 and -6 ..Which is true in both the cases..it shud be E..

Stem says that x is a positive number, thus x cannot be -6.

Re: If x is positive, is x > 3 [#permalink]
27 Oct 2013, 21:43

Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> (x+1)(x-3)>0 --> roots are -1 and 3.Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> (x+1)(x-5)>0 --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

Answer: D.

I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0

Re: If x is positive, is x > 3 [#permalink]
27 Oct 2013, 21:57

Expert's post

fozzzy wrote:

Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> (x+1)(x-3)>0 --> roots are -1 and 3.Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>3. Since given that x is positive then only one range is valid: x>3. Sufficient.

(2) (x - 2)^2 > 9 --> (x+1)(x-5)>0 --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: x<-1 or x>5. Since given that x is positive then only one range is valid: x>5. Sufficient.

Answer: D.

I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0