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a negative value of x

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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!
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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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