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

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

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New post 05 Nov 2014, 07:42
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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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New post 10 Nov 2014, 00:43
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\(a – b = √a - √b\)

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

\(\sqrt{a} = 1 - \sqrt{b}\)

\(a = 1 - 2\sqrt{b} + b\)

Answer = C
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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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New post 05 Nov 2014, 23:26
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(√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.
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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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New post 06 Nov 2014, 00:37
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sunaimshadmani wrote:
(√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
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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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New post 24 Jan 2016, 08:40
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?
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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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New post 24 Jan 2016, 21:25
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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.


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.

Answer (C)
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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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New post 25 Jan 2016, 07:07
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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New post 21 Mar 2016, 18:05
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?
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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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New post 21 Mar 2016, 21:29
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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New post 20 Jun 2017, 09:45
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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