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Math Expert V
Joined: 02 Sep 2009
Posts: 59586
If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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2
12 00:00

Difficulty:   65% (hard)

Question Stats: 56% (02:01) correct 44% (02:25) wrong based on 265 sessions

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Tough and Tricky questions: Algebra.

If a and b are different values and a – b = √a - √b, then in terms of b, a equals:

A. √b
B. b
C. b - 2√b + 1
D. b + 2√b + 1
E. b^2 – 2b√b + b

Kudos for a correct solution.

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Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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2
$$a – b = √a - √b$$

$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - {\sqrt{b}) = \sqrt{a} - {\sqrt{b}$$

$$\sqrt{a} = 1 - \sqrt{b}$$

$$a = 1 - 2\sqrt{b} + b$$

##### General Discussion
Manager  Joined: 05 Jun 2014
Posts: 60
GMAT 1: 630 Q42 V35 Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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2
(√a)^2 - (√b)^2 = √a - √b -----> ( √a + √b) ( √a - √b) = √a - √b -------> √a + √b = 1, so

√a = 1 - √b, square both sides and solve. Answer is C.
Intern  Joined: 06 May 2014
Posts: 4
Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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2
(√a)^2 - (√b)^2 = √a - √b -----> ( √a + √b) ( √a - √b) = √a - √b -------> √a + √b = 1, so

√a = 1 - √b, square both sides and solve. Answer is C.

Solution-
a-b=√a - √b
=> (√a + √b)(√a- √b)= √a- √b
so, either √a=√b
=> a=b

or, √a+√b=1
=> a=1+b-2√b

In my opinion, As b is not equal to a therefore, C should be answer
Intern  Joined: 03 Dec 2015
Posts: 1
Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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I'm having trouble understanding how the left side of the equation, when squared, isn't (a-b)(a-b)?

Why are we able to rewrite the left side as difference of squares, and how did we get radicals on the left side?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9850
Location: Pune, India
Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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3
Bunuel wrote:

Tough and Tricky questions: Algebra.

If a and b are different values and a – b = √a - √b, then in terms of b, a equals:

A. √b
B. b
C. b - 2√b + 1
D. b + 2√b + 1
E. b^2 – 2b√b + b

Kudos for a correct solution.

Here is another method:

We need a in terms of b. The options are in variable b. So what I instinctively want to do is plug in some values to get the answer.
a and b should be distinct.
$$a - b = \sqrt{a} - \sqrt{b}$$

So obviously, 1 comes to mind since $$1 = \sqrt{1}$$. But both a and b cannot be 1 since they must be distinct so I think of another value for which $$a = \sqrt{a}$$ and sure enough, it is 0.

If a = 0 and b = 1, then a – b = √a - √b (condition satisfied)
Put b = 1 in the options. Only options (C) and (E) give 0.

If a = 1 and b = 0, then again a – b = √a - √b (condition satisfied)
Put b = 0 in options (C) and (E); only option (C) gives 1.

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Karishma
Veritas Prep GMAT Instructor

Intern  Joined: 11 Oct 2012
Posts: 33
GMAT 1: 610 Q42 V32 Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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frankkn wrote:
I'm having trouble understanding how the left side of the equation, when squared, isn't (a-b)(a-b)?

Why are we able to rewrite the left side as difference of squares, and how did we get radicals on the left side?

Frankkn,
The left side uses the formula (a+b)(a-b)=a^2-b^2;
so , a - b has been rewritten as (√a+√b)(√a-√b).

Hope that helps !
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Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Algebra.

If a and b are different values and a – b = √a - √b, then in terms of b, a equals:

A. √b
B. b
C. b - 2√b + 1
D. b + 2√b + 1
E. b^2 – 2b√b + b

Kudos for a correct solution.

Bunuel - I don't think the question is worded correctly..should the a and b be INTEGERS, then the answer is C, but if not, then C is definitely not the answer.
if a and b are integers, then the difference between their roots is as well an integer.
we then have the first part the difference of 2 squares:
[sqrt(a) - sqrt(b)]*[sqrt(a)+sqrt(b) = sqrt(a) - sqrt(b)
we can conclude that sqrt(a)+sqrt(b) = 1
sqrt(a) = 1 - sqrt(b) - square both sides:
a = 1+b-2*sqrt(b)
or b - 2*sqrt(b) +1

C.

but I see that this works only if both a and b are integers and perfect squares...
anyone to explain why the "integer" part is not that important?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9850
Location: Pune, India
Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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frankkn wrote:
I'm having trouble understanding how the left side of the equation, when squared, isn't (a-b)(a-b)?

Why are we able to rewrite the left side as difference of squares, and how did we get radicals on the left side?

What is: $$(\sqrt{a} + \sqrt{b}) * (\sqrt{a} - \sqrt{b})$$?

Using the algebraic identity: $$(x + y)*(x - y) = x^2 - y^2$$, we get

$$(\sqrt{a} + \sqrt{b}) * (\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2$$

$$= a - b$$
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Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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IMO C
Indeed it is a simple yet tricky question.
(√a+√b)(√a-√b)=a^2-b^2
(√a+√b)=1
√a=1-√b........(1)
Squaring 1
We have b - 2√b + 1
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Re: If a and b are different values and a – b = √a - √b, then in terms of  [#permalink]

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_________________ Re: If a and b are different values and a – b = √a - √b, then in terms of   [#permalink] 11 Aug 2018, 10:08
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