If a – b

Question

If  a – b < 0  and  ab < 0 , what is  \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b| ?

A.  -2a  
 
B.  -a 
 
C.  ab 
 
D.  b  
 
E.  3b

MathRevolution · Accepted Answer

=> Since a – b < 0 and ab < 0 , we have b > 0 and a < 0. \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b| = \sqrt{(a-b)^2}+|a|-|b| = |a - b| + |a| - |b| = -(a - b) + (-a) – b since a – b < 0, a < 0, b > 0 = -a + b – a – b = -2a Therefore, A is the answer. Answer: A

nick1816 · Answer

ab<0 2 cases are possible 1. a>0 & b<0 but in this case a-b>0 (disregard this case) 2. a<0, b>0 a-b<0 in this case \sqrt{a^2-2ab+b^2} = |a-b|= b-a {as a-b<0} \sqrt{a^2}=|a|=-a {as a<0} |b| = b {b>0} \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b| = b-a+-a-b= -2a

Sheelarani · Answer

1. square root of (a^2 -2ab- b^2) can take 2 values:
- a-b
- b-a
because (a-b)^2 = (b-a)^2 and square root of (a-b)^2 can have both the possibilities.

Value 1.  a-b :
solution = a-b-a-b= -2b

Value 2.  b-a 
Solution= b-a-a-b= -2a

As answer choice A contains "-2a" that is why it is the correct answer.

chetan2u · Answer

a-b<0.......a<b 
 ab<0 , so a and b are of opposite sign.

But we know that a<b, so a should be negative and b positive.  a<0<b

We are looking for the value of  \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b|  
Let us solve each term separately. 
1)  \sqrt{a^2-2ab+b^2} = \sqrt{(a-b)^2} =|a-b|=b-a , as a-b<0
2)  \sqrt{a^2} = |a|=-a , as a<0
3)  |b|=b , as b>0
Substitute these values in  \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b| 
 |a-b|+|a|-|b|=b-a-a-b=-2a

A

Krishnahelps · Answer

MathRevolution   Bunuel  , Can't I take out a-b directly from root (a-b)^2? I already know that (a-b)^2 is positive.

shashankism · Answer

If a–b<0 , which means b>a.
and ab<0 which means wither a or b is -ve.

Since b>a , hence a is -ve and b is +ve
So, a<0, b>0
\sqrt{a^2−2ab+b^2} = \sqrt{(b-a)^2} = b-a (since square root is always +ve. b-a is +ve, a-b is -ve)
\sqrt(a^2} = -a (since square root is always +ve. a is -ve, -a is +ve
|b| = b (Since ,mod is always +ve, b is +ve, -b is -ve)

\sqrt{a^2−2ab+b^2} + \sqrt{a^2} -|b| = (b-a) + (-a) -b = -2a

Bunuel · Answer

(a - b)^2 is indeed nonnegative, as it's the square of a number. However, a - b could be negative, and since the square root cannot yield negative results,  \sqrt{a^2-2ab+b^2}  must equal |a - b|, specifically because the absolute value is also non-negative.

Next, we are given that a – b < 0, hence |a - b| = -(a - b). Therefore,  \sqrt{a^2-2ab+b^2} = |a - b| = -(a - b) .

As a general rule,  x^2=|x|^2 .

Hope this helps.

nivivacious · Answer

Easiest way to solve this as taking numbers. We know that a-b<0 so a<b and ab<0
so we can take a = -1 and b = 2
 \sqrt{a^2-2ab+b^2} + \sqrt{a^2} - |b| 
 \sqrt{(a-b)^2} + \sqrt{a^2} - |b| 
 \sqrt{(-1-2)^2} + \sqrt{-1^2} - |2| 
 \sqrt{-3^2} + \sqrt{1} - 2 
3+1-2
1+1
2
therefore 2a

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If a – b < 0 and ab < 0,

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