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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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If x=3/4 and y=1/5, what is the value of sqrt(x^2 + 6x + 9) [#permalink] New post 25 Jul 2007, 03:18
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

Please support your answer with explanations
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 [#permalink] New post 25 Jul 2007, 03:43
(B) for me :)

sqrt(x^2 + 6x + 9) - sqrt(y^2 - 2y + 1)
= sqrt((x+3)^2) - sqrt((y-1)^2)
= |x+3| - |y-1|
= |3/4 + 3| - |1/5-1|
= 15/4 - 4/5
= ((15*5) - 4*4) / 20
= (75-16)/20
= 59/20
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 [#permalink] New post 25 Jul 2007, 04:00
B for me as well. Fig's explanation is better than mine, but it works out to be (x+3)-(y+1) = (15/4)-(4/5) = (75/20)-(16/20) = 59/20
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 [#permalink] New post 25 Jul 2007, 06:01
OA is B.

Fig, you mean to say that for all such questions the rule applies..

E.g. sqrt((-3/5)^2) = 3/5 and not -3/5 ??

Am i getting it correct
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 [#permalink] New post 25 Jul 2007, 06:16
ajay_gmat wrote:
OA is B.

Fig, you mean to say that for all such questions the rule applies..

E.g. sqrt((-3/5)^2) = 3/5 and not -3/5 ??

Am i getting it correct


Yes :).... The result of a sqrt() must be positive or 0.

So, we cannot have for instance : sqrt(something) = - 4 (In the GMAT limits of course ;)).
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 [#permalink] New post 25 Jul 2007, 08:59
wow..Figs explanation is awsome..i did it the super long way...


but I get b after 2:30 minutes..Figs approach reduces this to less than 1 minute ...

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

well x^2=9/16 x=9/2

so 9/16+9/2+9 well

i multiplied it by 16/16 so I get; 9/16+(9.8)/16 + (9.16)/16

which reduced to (9(1+8+16))/16 or (9.25)/16 so sqrt[(9.25)/16]=3.5/4

same thing with y=1/25-2/5+1

multiply by 25/25 you get 1/25-10/25+25/25= (1-10+25)/25=16/25

sqrt(16/25)=4/5

15/4-4/5= 59/20

B :)

Gotta luv Figs approach..
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 [#permalink] New post 25 Jul 2007, 20:48
sqrt(9/16 + 18/4 + 9) - sqrt(1/25 - 2/5 + 1)
sqrt(225/16) - sqrt(16/25)
15/4 - 4/5 = 59/20
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 [#permalink] New post 27 Jul 2007, 10:01
Hi Fig,

Still having difficulty with thesqrt() being a positive result approach.

How do we then solve for x^2 = 1 ?

For solving this, we do Sqrt(x^2) = sqrt(1).. Hence here x should be only 1 and not x = -1,1 ??

Help.. :?
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 [#permalink] New post 27 Jul 2007, 23:44
ajay_gmat wrote:
Hi Fig,

Still having difficulty with thesqrt() being a positive result approach.

How do we then solve for x^2 = 1 ?

For solving this, we do Sqrt(x^2) = sqrt(1).. Hence here x should be only 1 and not x = -1,1 ??

Help.. :?


Have a look here on the whole approach to solve it :)

x^2 = 1
<=> Sqrt(x^2) = sqrt(1) >>>> we can use sqrt() to solve because x^2 >= 0 :)
<=> |x| = |1|
<=> |x| = 1

That implies
o If x > 0, then x = 1
or
o If x < 0, then -x = 1 <=> x = -1

Hope that helps :)
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 [#permalink] New post 28 Jul 2007, 01:59
Thanks Fig.. :-D

Your explanation helped.. The reasoning behind putting the absolute symbol || is to get a positive number always. There can be mutiple values for x.. But |x-1| will always be positive.. :-D
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 [#permalink] New post 28 Jul 2007, 04:41
ajay_gmat wrote:
Thanks Fig.. :-D

Your explanation helped.. The reasoning behind putting the absolute symbol || is to get a positive number always. There can be mutiple values for x.. But |x-1| will always be positive.. :-D


Note also that:

x^2 = 1
<=> x^2 - 1 = 0
<=> (x+1)*(x-1) = 0
<=> x=-1 or x=1

Another approach in order to avoid a deep undertsanding of what are the tricks behind sqrt(x) and x^2 functions ;)... Finally not a so good idea? ;)

In short :
o sqrt(x) = 1 => x = 1 as a the square root function is only definied for x >=0.
o x^2 = 1 => x=-1 or x=1 as the square function is definied for all x and is symetrical to the Y-axis ((-x)^2 = (-1)*(x)^2 = 1 * (x)^2 = x^2).
  [#permalink] 28 Jul 2007, 04:41
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