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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
5
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
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
1
This post was BOOKMARKED
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 _________________
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
Re: If x is positive, is x > 3 [#permalink]
01 Mar 2015, 19:47
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