If x^2-y = w, what is the value of x?
(1) w + y = 4
(2) y = 1
(1) w + y = 4
(2) y = 1
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Source: Beat The GMAT — Data Sufficiency |
i.e square x^2 - y = wGiorgio wrote:Sorry but what is that symbol in the middle?
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mihuna,maihuna wrote:i.e square x^2 - y = wGiorgio wrote:Sorry but what is that symbol in the middle?
is 2-y all to the power of x or only 2 ?
IMO C.
From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)
From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)
combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.
From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)
From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)
combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.
But When x=-2, (-2-1)(-2+1) = (-3)*(-1) = 3 = w so condition is still satisfied isn'theshamelaziry wrote:IMO C.
From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)
From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)
combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.
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Yu are correct ! still the answer should be C because we need to combine the statements to get the value of w. I am not sure why -2 works here ? it could be that the question is not well designed or so; this is not unusual with some DS questions. MAybe experts can help ?maihuna wrote:But When x=-2, (-2-1)(-2+1) = (-3)*(-1) = 3 = w so condition is still satisfied isn'theshamelaziry wrote:IMO C.
From 1: x^2 = w + y = 4 --> x = -2 or x = 2 (insuff)
From 2: x^-1 = w---> (x-1)(x+1) = w (insuff)
combined: w = 3 ---> put 3 in statement 2 and solve for x we get x =2 or x =4 . Here we take x =2 as a solution as it conforms to the same x value we got from statement 1.
You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!
You just get X=+2 or -2 , since equation is X^2-1=3
So answer is E !
You just get X=+2 or -2 , since equation is X^2-1=3
So answer is E !
Algebraically, why can't I do x-1=3 and x+1 = 3 ???????? if instead of 3 there were a zero, we do this. What is the difference ?Giorgio wrote:You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!
You just get X=+2 or -2 , since equation is X^2-1=3
So answer is E !
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The difference is the special properties of the number 0.heshamelaziry wrote:Algebraically, why can't I do x-1=3 and x+1 = 3 ???????? if instead of 3 there were a zero, we do this. What is the difference ?Giorgio wrote:You can not solve 2nd equation like that! (x-1)(x+1)=3 does not give you X=2 and x=4 ! 3 is not 0!
You just get X=+2 or -2 , since equation is X^2-1=3
So answer is E !
The ONLY way to get a product of 0 is to multiply by 0. There are infinite ways to get a product of 3.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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Hesham,
Let's say we have (x-5)(x+7) = 0
See that this is just saying:
(one number)(another number) = 0
Or
x*y = 0
The only way for x*y = 0 to be true is if at least one of x or y equals 0.
This is why (x-5)(x+7) = 0 necessitates that either x = 5 or x = -7.
But when you have a non-zero number there will be a whole bunch of numbers whose product will be that number; in fact, if you consider non-integers, there are an infinite pairs of numbers whose product is 3.
Let's say we have (x-5)(x+7) = 0
See that this is just saying:
(one number)(another number) = 0
Or
x*y = 0
The only way for x*y = 0 to be true is if at least one of x or y equals 0.
This is why (x-5)(x+7) = 0 necessitates that either x = 5 or x = -7.
But when you have a non-zero number there will be a whole bunch of numbers whose product will be that number; in fact, if you consider non-integers, there are an infinite pairs of numbers whose product is 3.
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