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by rahul.s » Thu Feb 11, 2010 2:36 am
if x is not = 0, then sqrt x^2/x = ?

(A) -1
(B) 0
(C) 1
(D) x
(E) |x| / x

i opted for C, but the OA is E

didn't quite understand why.
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by sanjayism » Thu Feb 11, 2010 2:48 am
d) is correct answer

x^2/x =x
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by papgust » Thu Feb 11, 2010 3:17 am
Here x can both -ve and +ve

sqrt (x^2) = +/- x = |x|

So, sqrt (x^2) / x = |x| / x (Choice E)
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by thephoenix » Thu Feb 11, 2010 3:31 am
rahul.s wrote:if x is not = 0, then sqrt x^2/x = ?

(A) -1
(B) 0
(C) 1
(D) x
(E) |x| / x

i opted for C, but the OA is E

didn't quite understand why.
x^2 ---> x can be -ve or +ve---->lxl

so x^2/x=lxl/x

another way is substitue x as 1,-1
for x=1; x^2/x=1
for x=-1;x^2/x=-1

which is smae as E
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by mgmt_gmat » Thu Feb 11, 2010 4:59 am
HERE we don't know x>0 or x< 0. Hence sqrt(square of x) is |x|. that means +x or -x.
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by shashank.ism » Mon Feb 15, 2010 12:17 pm
rahul.s wrote:if x is not = 0, then sqrt x^2/x = ?

(A) -1
(B) 0
(C) 1
(D) x
(E) |x| / x

i opted for C, but the OA is E

didn't quite understand why.
rahul if x is not 0 so x may be either positive or negative....
hence sqrt.(x^2) = |x| = x for x>0
and -x for x<0.
so, sqrt.(x^2)/x =[spoiler] |x|/x Ans E[/spoiler]
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