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If \(x = \frac{3}{4}\) and \(y = \frac{2}{5}\), what is the value of \(\sqrt{(x^2 + 6x + 9)} - \sqrt{(y^2 - 2y +1)}\)?

A. \(\frac{87}{20}\)
B. \(\frac{63}{20}\)
C. \(\frac{47}{20}\)
D. \(\frac{15}{4}\)
E. \(\frac{14}{5}\)
Spoiler: OA

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If \(x = \frac{3}{4}\) and \(y = \frac{2}{5}\), what is the value of \(\sqrt{(x^2 + 6x + 9)} - \sqrt{(y^2 - 2y +1)}\)?

A. \(\frac{87}{20}\)
B. \(\frac{63}{20}\)
C. \(\frac{47}{20}\)
D. \(\frac{15}{4}\)
E. \(\frac{14}{5}\)

\(\sqrt{(x+3)^2} - \sqrt{(y-1)^2} = |x+3| - |y-1| = | \frac{3}{4} + 3| - |\frac{2}{5}-1| = \frac{15}{4}-\frac{3}{5}=\frac{63}{20}\)


Answer: B
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Why do we use absolute signs in this problem??
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rsamant wrote:
Why do we use absolute signs in this problem??


Because \(\sqrt{x^2}=|x|\). Check for more Absolute value tips: absolute-value-tips-and-hints-175002.html


Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

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Re: M06-02 [#permalink]

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New post 05 Jan 2015, 02:18
Hi Bunuel,
Agreed that as per the theory & property its \(\sqrt{x^2}=|x|\)
But why can't we cancel out the square root with the power square?
Because i did it as \(\sqrt{(x+3)^2} - \sqrt{(y-1)^2}\)
after cancelling the roots ----> (x+3)-(y-1)---> \(x-y+4 = 3/4 - 2/5 + 4\)
and answer came out to be 87/20.
Please help
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Re: M06-02 [#permalink]

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Ankur9 wrote:
Hi Bunuel,
Agreed that as per the theory & property its \(\sqrt{x^2}=|x|\)
But why can't we cancel out the square root with the power square?
Because i did it as \(\sqrt{(x+3)^2} - \sqrt{(y-1)^2}\)
after cancelling the roots ----> (x+3)-(y-1)---> \(x-y+4 = 3/4 - 2/5 + 4\)
and answer came out to be 87/20.
Please help


Because \(\sqrt{(x+3)^2}=|x+3|\), not x+3 and \(\sqrt{(y-1)^2} = |y-1|\), not y-1. Check for more here: absolute-value-tips-and-hints-175002.html
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Re: M06-02 [#permalink]

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New post 05 Jan 2015, 06:02
Bunuel wrote:
Ankur9 wrote:
Hi Bunuel,
Agreed that as per the theory & property its \(\sqrt{x^2}=|x|\)
But why can't we cancel out the square root with the power square?
Because i did it as \(\sqrt{(x+3)^2} - \sqrt{(y-1)^2}\)
after cancelling the roots ----> (x+3)-(y-1)---> \(x-y+4 = 3/4 - 2/5 + 4\)
and answer came out to be 87/20.
Please help


Because \(\sqrt{(x+3)^2}=|x+3|\), not x+3 and \(\sqrt{(y-1)^2} = |y-1|\), not y-1. Check for more here: absolute-value-tips-and-hints-175002.html


I checked the absolute tips and \(\sqrt{x^2}=|x|\) is given as one of the properties
But how the property is derived or holds is not mentioned.
Is this property holds because in GMAT only positive values are considered for even roots and thus \(\sqrt{x^2}=|x|\)
right??
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Ankur9 wrote:
Bunuel wrote:
Ankur9 wrote:
Hi Bunuel,
Agreed that as per the theory & property its \(\sqrt{x^2}=|x|\)
But why can't we cancel out the square root with the power square?
Because i did it as \(\sqrt{(x+3)^2} - \sqrt{(y-1)^2}\)
after cancelling the roots ----> (x+3)-(y-1)---> \(x-y+4 = 3/4 - 2/5 + 4\)
and answer came out to be 87/20.
Please help


Because \(\sqrt{(x+3)^2}=|x+3|\), not x+3 and \(\sqrt{(y-1)^2} = |y-1|\), not y-1. Check for more here: absolute-value-tips-and-hints-175002.html


I checked the absolute tips and \(\sqrt{x^2}=|x|\) is given as one of the properties
But how the property is derived or holds is not mentioned.
Is this property holds because in GMAT only positive values are considered for even roots and thus \(\sqrt{x^2}=|x|\)
right??


About \(\sqrt{x^2}=|x|\):

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).
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New post 14 Oct 2015, 11:33
Hi Bunuel,
I understand your explanation.
I wanted to apply the mod and solve this problem, However, we also know that..."When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
So, the solution to this problem can go either way 87/20 (considering the positive roots only) or 63/20 ( with the mod approach) ...
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rohitd80 wrote:
Hi Bunuel,
I understand your explanation.
I wanted to apply the mod and solve this problem, However, we also know that..."When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root".
So, the solution to this problem can go either way 87/20 (considering the positive roots only) or 63/20 ( with the mod approach) ...



PEMDAS
parentheses comes first. we cannot jump operations and square/take the square root of everything before solving for what is in parenthesis.
y-1 results in a negative expression. After squaring it, it gives us a positive expression. Moreover, after squaring everything, the square root will be as well a positive number. I did not do the way bunuel explained, yet I got to the right answer.

sqrt[(x+3)^2] = sqrt[(3/4+12/4)^2] = sqrt[(15/4)^2]. this will result in 15/4.
now let's take y
sqrt[(y-1)^2] = sqrt[(2/5-5/5)^2] = sqrt[(-3/5)^2]. now, if we square -3/5 we get 9/25 which is a positive number. sqrt of 9/25 is 3/5.

we got 15/4 - 3/5
multiply first by 5/5 and second by 4/4, the result is:
(15*5)/20 - (3*4)/20. extend this and get: (75-12)/20.
the result is 63/20.
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New post 14 Dec 2015, 14:39
Hello Bunuel,
As per Official Guide, |x| is defined to be x if x >= 0 and -x if x < 0. Here it is |x+3| - |y-1| = |15/4| - |-3/5|.
Now (y-1) = -3/5 which is less than 0.

So shouldn't it be |15/4| - |-3/5| = (15/4) - (-3/5) = (15/4) + (3/5) = 87/20 ?

Kindly correct me if I'm making a mistake. Thanks.
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Devlikes wrote:
Hello Bunuel,
As per Official Guide, |x| is defined to be x if x >= 0 and -x if x < 0. Here it is |x+3| - |y-1| = |15/4| - |-3/5|.
Now (y-1) = -3/5 which is less than 0.

So shouldn't it be |15/4| - |-3/5| = (15/4) - (-3/5) = (15/4) + (3/5) = 87/20 ?

Kindly correct me if I'm making a mistake. Thanks.


|x| = -x if x <= 0, yes. But the point behind this is that absolute value cannot be negative, for example |-5| = 5. If x is negative, then |x| = -x = -negative = positive. Here, \(|-\frac{3}{5}| = -(-\frac{3}{5}) = \frac{3}{5}\)

Hope it's clear.
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New post 28 Jul 2016, 10:43
When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root . Thats why i didn't use mod. What am i missing here?
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New post 29 Jul 2016, 04:13
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born222 wrote:
When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root . Thats why i didn't use mod. What am i missing here?


Precisely because the square root sign cannot give negative result you should use modulus sign here. x + 3 depending on x can be negative, while |x + 3| cannot.

Check for more the discussion above, specifically the following post: m06-183699.html#p1465193

Hope it helps.
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New post 22 Aug 2016, 11:19
I think this is a high-quality question and I agree with explanation. I think you can make this question a bit harder if you add 83/20 as an answer choice for people like me who forgot to take the absolute value into the consideration.
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New post 25 Jul 2017, 19:46
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.


Please read the discussion above and ask specific question.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation. Explanation is fine. It's only missing a "+" sign between 3 and 3/4
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New post 11 Feb 2018, 21:45
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rodmm wrote:
I think this is a high-quality question and I agree with explanation. Explanation is fine. It's only missing a "+" sign between 3 and 3/4


Thank you. Edited the typo.
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Re: M06-02   [#permalink] 11 Feb 2018, 21:45
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