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

Question

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

PareshGmat · Accepted Answer

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

KarishmaB · Answer

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)

sunaimshadmani · Answer

(√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.

vinit9211 · Answer

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

frankkn · Answer

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?

AndyNeedsGMAT · Answer

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 !

mvictor · Answer

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)
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?

KarishmaB · Answer

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

arvind910619 · Answer

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

BrentGMATPrepNow · Answer

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

LLM1500 · Answer

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???

kajaldaryani46 · Answer

—-
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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Vinayak1996 · Answer

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

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Bunuel · Answer

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

georgia123 · Answer

I'm still really confused on this part ... can someone please break it down further? i feel like the answer is staring me in the face and i'm just not grasping the concept

Bunuel · Answer

(\sqrt{a})^2 =\sqrt{a}*\sqrt{a}= a

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

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