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If a – b < 0 and ab < 0,

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MathRevolution
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If a – b < 0 and ab < 0, [#permalink] New post   06 Mar 2020, 03:06
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[GMAT math practice 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. \(3b\)
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A
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Re: If a – b < 0 and ab < 0, [#permalink] New post   08 Mar 2020, 20:07
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=>

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
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Re: If a – b < 0 and ab < 0, [#permalink] New post   06 Mar 2020, 04:02
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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


MathRevolution wrote:
[GMAT math practice 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. \(3b\)
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Re: If a – b < 0 and ab < 0, [#permalink] New post   18 Apr 2021, 04:29
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.


nick1816 wrote:
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


MathRevolution wrote:
[GMAT math practice 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. \(3b\)
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Re: If a – b < 0 and ab < 0, [#permalink] New post   18 Apr 2021, 05:11
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MathRevolution wrote:
[GMAT math practice 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. \(3b\)



\(a-b<0.......a<b\)
\(ab<0\), 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
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Re: If a b < 0 and ab < 0, [#permalink] New post   01 Aug 2023, 22:21
MathRevolution wrote:
=>

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


MathRevolution Bunuel , Can't I take out a-b directly from root (a-b)^2? I already know that (a-b)^2 is positive.
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Re: If a b < 0 and ab < 0, [#permalink] New post   01 Aug 2023, 23:02
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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
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Re: If a b < 0 and ab < 0, [#permalink] New post   02 Aug 2023, 00:56
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ankitapugalia wrote:
MathRevolution wrote:
=>

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


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


(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.
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Re: If a b < 0 and ab < 0, [#permalink] New post   24 Oct 2023, 22:37
MathRevolution wrote:
[GMAT math practice 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. \(3b\)


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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Re: If a b < 0 and ab < 0, [#permalink]
24 Oct 2023, 22:37
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