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a negative value of x

by sanju09 » Fri Jun 18, 2010 4:09 am
For which of a negative value of x the relation [(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 is true?
(A) -5
(B) -4
(C) -3
(D) -2
(E) -1
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by kvcpk » Fri Jun 18, 2010 4:12 am
[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0
If you observe closely, internal bracket turns x^2
so (x^2)^1/2 + 2|X| - 6 = 0

(x^2)^1/2 = x [not -x]

x+2|X| - 6 = 0

if x>0 then |X| = x else -x
assume x>0, then
3x-6 = 0 which gives x = 2

when x<0
-x - 6 = 0
x= -6

question is asking us for negative value so Answer -6
Last edited by kvcpk on Fri Jun 18, 2010 4:19 am, edited 2 times in total.
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by amising6 » Fri Jun 18, 2010 4:15 am
For which of a negative value of x the relation [(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 is true?
(A) -5
(B) -4
(C) -3
(D) -2
(E) -1

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0
[x^2+2x+1 -2x-1]^1/2 + 2|x| - 6 = 0
x+ 2|x| - 6 = 0
this can be written as
x+2x-6=0 x=2
and x-2x-6=0 x=-6

so ans wer will be -6
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by singhpreet1 » Sat Jun 19, 2010 12:35 am
amising6 wrote:For which of a negative value of x the relation [(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 is true?
(A) -5
(B) -4
(C) -3
(D) -2
(E) -1

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0
[x^2+2x+1 -2x-1]^1/2 + 2|x| - 6 = 0
x+ 2|x| - 6 = 0
this can be written as
x+2x-6=0 x=2
and x-2x-6=0 x=-6

so ans wer will be -6
how can you put

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 under the same exponential power..what property am i missing here?

(x+1)^2 - (2x+1)^1/2 does not share the same exponent does it?


thanks. Preet.
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by sanju09 » Sat Jun 19, 2010 12:44 am
singhpreet1 wrote:
amising6 wrote:For which of a negative value of x the relation [(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 is true?
(A) -5
(B) -4
(C) -3
(D) -2
(E) -1

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0
[x^2+2x+1 -2x-1]^1/2 + 2|x| - 6 = 0
x+ 2|x| - 6 = 0
this can be written as
x+2x-6=0 x=2
and x-2x-6=0 x=-6

so ans wer will be -6
how can you put

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 under the same exponential power..what property am i missing here?

(x+1)^2 - (2x+1)^1/2 does not share the same exponent does it?


thanks. Preet.
Hi Preet, your doubt is most welcome. Actually, the exponent ½ is on the entire [(x + 1) ^2 - (2 x + 1)], which reduces to x^2 bearing an exponent ½, which is finally x. Rest is already very well explained by amising6 above. It could be the bracket's property that you were missing here. Stay tuned!
The mind is everything. What you think you become. -Lord Buddha



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by singhpreet1 » Sat Jun 19, 2010 12:52 am
yea got that..right on Sanju...my bad!
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by amising6 » Sat Jun 19, 2010 12:56 am
singhpreet1 wrote:
amising6 wrote:For which of a negative value of x the relation [(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 is true?
(A) -5
(B) -4
(C) -3
(D) -2
(E) -1

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0
[x^2+2x+1 -2x-1]^1/2 + 2|x| - 6 = 0
x+ 2|x| - 6 = 0
this can be written as
x+2x-6=0 x=2
and x-2x-6=0 x=-6

so ans wer will be -6
how can you put

[(x + 1) ^2 - (2 x + 1)] ^½ + 2|x| - 6 = 0 under the same exponential power..what property am i missing here?

(x+1)^2 - (2x+1)^1/2 does not share the same exponent does it?


thanks. Preet.

[(x + 1) ^2 - (2 x + 1)] ^½


[(x + 1) ^2 - (2 x + 1)] this whole is raised to power 1/2
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by mj78ind » Sat Jun 19, 2010 1:02 am
sqrt((x+1)^2-(2x+1)) + 2abs(x) - 6 =0

gives abs(x) + 2abs(x) - 6 =0

which is 3abs(x) = 6 or abs(x) = 2

Hence x = -2

Answer D

OA pls?
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