Square Root

IMO B

Ques: sqrt[(x-3)^2] = (3-x)?

sqrt[(x-3)^2] = (x-3) OR (3-x)

St 1: x is not equal to 3

If x > 3 then the solution will be equal to (x-3)

Take x=4, sqrt(4-1)^2 = 1 and (4-3) = 1

If x<3 then the solution will be equal to (3-x)

Take x=2, sqrt(2-3)^2 = 1 and (3-2) = 1 [(x-3) will give you a negative value].

Hence St1 is INSUFF.

St2 -xlxl > 0

lxl is always positive

for -x to be positive, x has to be negative.

Since x is negative x<3 , the equation will have only one solution (3-x) [refer the explanation for x=2 in st1].

So B

sunnyguest wrote:Hi,

Can anyone help me out on the DS problem below?

Is [(x-3)^2]^0.5 = 3-x ?

1) x /= 3

2) -xlxl > 0

solving stem x-3= 3-x,hence X=3

meaning LHS=RHS only for X=3

1. X is not equal to 3

means LHS is not qual to RHS

Its a yes or no question hence 1 is sufficient

2.

Lets say X=3

-3*abs(3)>0

-3*3>0

-9>0

Not true hence X not equal to 3

X=4, -12>0 not true

X=-3, 9>0 True

hence, in any case condition is satisifed only for negative nos

sure that X is not equal to 3,means LHS is not equal to RHS

Its a yes or no question hence 2 is sufficient


D

Iamcaste, since x is not given as a positive number negative intgers are fair game

One more vote for B)

Stmt I

Substitute for x in the eqn i.e

Take x=-3 - YES
Take x = 4 - NO

INSUFF

Stmt II

X has to be negative for -x|x| > 0


x=-3
x=-5
.....etc...
Provides YES always i.e LHS = RHS

B)

Another for B.

Same approach as cramya

Yes, its B....-3 can also be a possibility..