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If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9)

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ajay_gmat
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If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   Updated on: 25 Jul 2007, 02:52
If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) - sqrt(y^2 - 2y + 1) ?

(a)91/20
(b)59/20
(c)47/20
(d)15/4
(e)14/5

So sorry, I had to edit my Question. Please support your answers with explanations.



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Originally posted by ajay_gmat on 24 Jul 2007, 22:46.
Last edited by ajay_gmat on 25 Jul 2007, 02:52, edited 1 time in total.
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   24 Jul 2007, 23:26
ajay_gmat wrote:
If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) - sqrt(y^2 - 2y + 1) ?

(a)89/20
(b)59/20
(c)47/20
(d)15/4
(e)14/5


Answer B

first sqrt solves to sqrt (x+3)^2, second sqrt solves to sqrt (y-1)^2

Plug in values of x and y to get 15/4-4/5 which is equal to 59/20
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 01:38
GK_Gmat wrote:
ajay_gmat wrote:
If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) - sqrt(y^2 - 2y + 1) ?

(a)89/20
(b)59/20
(c)47/20
(d)15/4
(e)14/5


Answer B

first sqrt solves to sqrt (x+3)^2, second sqrt solves to sqrt (y-1)^2

Plug in values of x and y to get 15/4-4/5 which is equal to 59/20


I think A

But

x + 3 - (y-1) = x -y + 4 = 4.55

15/4 -(-4/5) =

75 + 16
---------
5

But my problem is 4.55 is not a choice there... even 89/20 is only 4.45 if I am right.

Am I missing something?
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 01:41
I get it

sqrt can be +/-

3.75 -[ -(y-1)] is also a possible answer.

So 59/20 makes sense.
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 10:21
59/20

Got to be careful with squares and roots ..
i was about to select 91/20 and when u mentioned be careful :)
thought twice about the answer.
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 11:37
got A

sqrt [x^2+6x+9] = sqrt[(x+3)(x+3)] = x+3 = (3/4)+3 or 15/4
sqrt[y^2-2y+1] = sqrt[(y-1)(y-1)] = y-1 = (1/5)-1 or -4/5

so (15/4) - (-(4/5)) = (15/4) + (4/5) = (75/20) + (16/20) =91/20.

what am i missing that people are getting (59/20)?

what's the OA?
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 12:16
I get B. 59/20

Plugging in values I get 15/4 - 4/5 which equals to 59/20

I think the trick is knowing that sqrt(y-1)^2 must be solved inside the parentheses first, before squaring the inside, then taking its square root. If you just have the powers cancel first, you would get 15/4 - (-4/5), thus equaling 91/20 the wrong answer.
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 15:47
emoryhopeful wrote:
I get B. 59/20

Plugging in values I get 15/4 - 4/5 which equals to 59/20

I think the trick is knowing that sqrt(y-1)^2 must be solved inside the parentheses first, before squaring the inside, then taking its square root. If you just have the powers cancel first, you would get 15/4 - (-4/5), thus equaling 91/20 the wrong answer.


Btw, I dont think 91/20 is a wrong answer... Its just not one of the answer choices here. So we were left to find one of more possible right answers.
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 16:12
after reading it, I worked it out and got A, little fuzzy on how folks are getting B.

isn't it 75/20 - (-16/20) ?
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 16:18
zakk wrote:
after reading it, I worked it out and got A, little fuzzy on how folks are getting B.

isn't it 75/20 - (-16/20) ?



I guess they are saying that you can't have a negative root.

What's the source of this question?
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 16:25
Hayabusa wrote:
zakk wrote:
after reading it, I worked it out and got A, little fuzzy on how folks are getting B.

isn't it 75/20 - (-16/20) ?



I guess they are saying that you can't have a negative root.

What's the source of this question?


Right. But if we have handled the sqrt by eliminating the ^2 (they cancel don't they or am I really off!?) doesn't it come a moot point?
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 16:53
ahhhh, it clicked.

It's because you are plugging in the numbers for the variables and squaring them BEFORE taking the square root.

That makes sense, I get it and am now a believer in option B.

Good question.

Can we confirm the OA???
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   25 Jul 2007, 23:20
Other approaches here :

https://www.gmatclub.com/phpbb/viewtopic.php?t=49388

The OA is indeed B :)

(ajay_gmat : pls, do not post 2 times the same question... thx :))
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Re: If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post   26 Jul 2007, 09:24
Hayabusa wrote:
zakk wrote:
after reading it, I worked it out and got A, little fuzzy on how folks are getting B.

isn't it 75/20 - (-16/20) ?



I guess they are saying that you can't have a negative root.

What's the source of this question?


I remember reading something in Manhattan GMAT that said that for GMAT purposes, the square root is always the positive square root.



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