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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   95% (hard)

Question Stats: 44% (02:06) correct 56% (01:57) wrong based on 129 sessions

### HideShow timer Statistics 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. $$\frac{14}{5}$$

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Math Expert V
Joined: 02 Sep 2009
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Official Solution:

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. $$\frac{14}{5}$$

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

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Why do we use absolute signs in this problem??
Math Expert V
Joined: 02 Sep 2009
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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

Hope this helps.
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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.
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Math Expert V
Joined: 02 Sep 2009
Posts: 56300

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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.

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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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.

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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Math Expert V
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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.

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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Hi Bunuel,
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 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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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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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?
Math Expert V
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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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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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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
Math Expert V
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VeteranMom wrote:
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 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
Math Expert V
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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.
_________________ Re: M06-02   [#permalink] 11 Feb 2018, 21:45

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# M06-02

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