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# If a and b are constants, is the expression

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If a and b are constants, is the expression  [#permalink]

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27 Apr 2019, 12:25
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If a and b are constants, is the expression $$\frac{x+b}{\sqrt{x+a}}$$ defined for x = –2 ?

1) a = 5
2) b = 6

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Re: If a and b are constants, is the expression  [#permalink]

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27 Apr 2019, 12:43
3
carcass wrote:
If a and b are constants, is the expression $$\frac{x+b}{\sqrt{x+a}}$$ defined for x = –2 ?

1) a = 5
2) b = 6

OG2020 NEW QUESTION
DS05330

Value of expression depends entirely onvalue of a
since a is what makes the expression either infinite, non real or real

$$\frac{x+b}{\sqrt{x+a}}$$

Stmt one tells a =5
which makes denominator real
and value of expression becomes
$$\frac{-2+b}{\sqrt{-2+5}}$$ = $$\frac{-2+b}{\sqrt{3}}$$

hence sufficient

Stmt II gives no info for a
hence insufficient
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Re: If a and b are constants, is the expression  [#permalink]

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29 Apr 2019, 10:49
Necessary condition is (1) as allows to define the numerator (making sure no negative root arises). The denominator is already defined no matter what. Even if b=2, the whole fraction would result in 0 which is defined.
Condition (2) isn't useful as does not tell us anything about the denominator. Denominator could still be root of a negative number.
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If a and b are constants, is the expression  [#permalink]

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Updated on: 16 Jan 2020, 09:57
2
Top Contributor
2
carcass wrote:
If a and b are constants, is the expression $$\frac{x+b}{\sqrt{x+a}}$$ defined for x = –2 ?

1) a = 5
2) b = 6

Target question: Is the expression defined for x = –2?
This is a great candidate for rephrasing the target question.

If x = -2, then the expression becomes $$\frac{(-2)+b}{\sqrt{(-2)+a}}$$

There are two ways in which the expression $$\frac{(-2)+b}{\sqrt{(-2)+a}}$$ is NOT defined:

case i) If a = 2, then the fraction's denominator is 0, which means the entire rational expression is NOT defined.
case ii) If a < 2, then the value inside the square root sign is NEGATIVE, in which case the denominator is NOT defined.
When we combine both cases, we see that the expression is NOT defined when a ≤ 2. ..
Or we can say, the expression IS defined when a > 2

REPHRASED target question: Is a > 2?

Aside: the video below has tips on rephrasing the target question

Statement 1: a = 5
The answer to the REPHRASED target question is YES, a IS greater than 2
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: b = 6
This does not help us answer the REPHRASED target question.
Statement 2 is NOT SUFFICIENT

Cheers,
Brent

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Originally posted by GMATPrepNow on 29 Apr 2019, 11:51.
Last edited by GMATPrepNow on 16 Jan 2020, 09:57, edited 1 time in total.
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Re: If a and b are constants, is the expression  [#permalink]

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09 May 2019, 17:27
1
Hi All,

We're told that A and B are constants. We're asked if the expression (X+B)/(√(X+A)) is "defined" for x = -2. This is a YES/NO question and is a great 'concept' question - meaning that you don't have to do much math if you understand the concepts involved. For the given fraction to be 'defined', the denominator CANNOT be 0. As such, since we're given a value for X (re: -2), we need to know the value for A to answer this question.

1) A = 5

Fact 1 gives us the value of A. With the given value of X, we know that the denominator will be a POSITIVE number, so the fraction will be defined and the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

2) B = 6

Fact 2 gives us information to define the numerator of the fraction (re: -2 + 6 = +4), but not the denominator.
IF....
A = +2, then the answer to the question is NO
A = +3, then the answer to the question is YES
Fact 2 is INSUFFICIENT

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If a and b are constants, is the expression  [#permalink]

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Updated on: 23 May 2019, 00:55
Can some one please explain what ''defined'' mean in this context or rephrase this question in an easier way?

Does defined mean terminating? Does the question mean whether this fraction becomes a terminating decimal?

Originally posted by rebinh1982 on 22 May 2019, 04:48.
Last edited by rebinh1982 on 23 May 2019, 00:55, edited 2 times in total.
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Re: If a and b are constants, is the expression  [#permalink]

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22 May 2019, 10:00
2
rebinh1982 wrote:
Can some one please explaine what ''defined'' mean in this context or rephrase this question in an easier way?

Does define mean terminating? Does the question mean whether this fraction becomes a terminating decimal?

Hi rebinh1982,

For the given fraction to be 'defined', the denominator CANNOT be 0. In addition, since we have a square-root in the denominator, we also cannot have a negative total (since that would create an imaginary number). As such, since we're given a value for X (re: -2), we need to know the value for A to answer this question. In simple terms, if A is less than or equal to 2, then the answer to the question is NO. If A is greater than 2, then the answer to the question is YES.

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Re: If a and b are constants, is the expression  [#permalink]

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23 May 2019, 00:50
EMPOWERgmatRichC

Thanks a lot for your explanation, now i understand the question 100%!
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If a and b are constants, is the expression  [#permalink]

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16 Jan 2020, 09:50
1
GMATPrepNow wrote:
If a and b are constants, is the expression $$\frac{x+b}{\sqrt{x+a}}$$ defined for x = –2 ?

1) a = 5
2) b = 6

REPHRASED target question: Is a > –2?

GMATPrepNow shouldn't it be Is a > 2?

bcoz for a = -1 the root still remains -ve
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Re: If a and b are constants, is the expression  [#permalink]

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16 Jan 2020, 09:58
Top Contributor
Neena96 wrote:
GMATPrepNow wrote:
If a and b are constants, is the expression $$\frac{x+b}{\sqrt{x+a}}$$ defined for x = –2 ?

1) a = 5
2) b = 6

REPHRASED target question: Is a > –2?

GMATPrepNow shouldn't it be Is a > 2?

bcoz for a = -1 the root still remains -ve

Good catch - thanks!!
In my solution, 2 somehow morphed into -2
I have edited my response accordingly.

Kudos for you!!!!

Cheers,
Brent
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Re: If a and b are constants, is the expression   [#permalink] 16 Jan 2020, 09:58
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