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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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(√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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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]
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]
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]
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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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]
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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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.


Key concept: \((x^2 - y^2) = (x + y)(x - y)\) AKA, factoring a difference of squares

Now all we need to do is recognize that a = (√a)² and b = (√b)²

So, we can now take: a – b = √a - √b
Rewrite the left side as follows: (√a)² – (√b)² = √a - √b
Factor the left side: (√a + √b)(√a - √b) = √a - √b
Divide both sides by (√a - √b) to get: (√a + √b) = 1
Subtract √b from both sides to get: √a = 1 - √b
Square both sides to get: (√a)² = (1 - √b)²
In other words: (√a)² = (1 - √b)(1 - √b)
Expand and simplify both sides: a = 1 - 2√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]
How can you divide by (Square root A - Square root B) to get 1 on the right hand side of the equation? I thought that when you divide you always need to know whether its a positive or negative term. We would have no way of knowing whether that term is positive, thus equaling positive 1 on the right hand side of the equation or whether the term is negative, thus equaling negative 1 on the right hand side of the equation???
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Re: If a and b are different values and a b = a - b, then in terms of [#permalink]
VeritasKarishma wrote:
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)

—-
I used this method by picking a = 1 and b=0 and got down to two options C and D - but didn’t know how to eliminate further. So I just guessed C but per your method, we should verify the value of not just a (whether that’s equal to 1) but also whether the value of b is 1 when when a=0 right?

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Re: If a and b are different values and a b = a - b, then in terms of [#permalink]
Why is B wrong.
When we substitute a=b then we get LHS = RHS
Then why is it still wrong?

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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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Vinayak1996 wrote:
Why is B wrong.
When we substitute a=b then we get LHS = RHS
Then why is it still wrong?

Posted from my mobile device


Check the highlighted part in the question stem:
If a and b are different values...
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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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