sqrt refers to the positive root
sqrt ((x-3)^2) can yield either x-3 or 3-x,whichever one is positive

2 tells us that x is negative
3-x is positive and thus the root.

So 2 is clearly sufficient

1 is not sufficient


B is the answer for me!

The Q stem can be simplefied to say x-3 = 3-x ie 2x=6 or is x=3

1 is sufficient

2 says x is negative so cannot be equal to 3 hence sufficient

so ans is D

RAGS wrote:The Q stem can be simplefied to say x-3 = 3-x ie 2x=6 or is x=3

wrong!

sqrt(x-3)^2 is x-3 when x>3 and 3-x when x<3

ajith wrote:
sqrt(x-3)^2 is x-3 when x>3 and 3-x when x<3

Ok, this is the part I don't understand.

Take x = -1, (-1-3)^2 = 16 and root-16 can be +4 or -4 right ?

Why should the root operation yield only the positive root ?

jayhawk2001 wrote:

ajith wrote:
sqrt(x-3)^2 is x-3 when x>3 and 3-x when x<3

Ok, this is the part I don't understand.

Take x = -1, (-1-3)^2 = 16 and root-16 can be +4 or -4 right ?

Why should the root operation yield only the positive root ?

A square root, also called a radical or surd, of x is a number r such that r^2==x. The function sqrt(x) is therefore the inverse function of f(x)==x^2 for x>=0.
Source :https://mathworld.wolfram.com/SquareRoot.html


Please refer the attached image

ajith wrote: A square root, also called a radical or surd, of x is a number r such that r^2==x. The function sqrt(x) is therefore the inverse function of f(x)==x^2 for x>=0.
Source :https://mathworld.wolfram.com/SquareRoot.html

Ah, ok. Official Guide Quantitative Review (Green book) page 24,
section 7, also says the following --

"Ever positive number n has two square roots, one positive and the
other negative, but root-n denotes the positive number whose square
is n".

Darn, missed this.

Just to summarize, x^2 = 9 implies x can be +3 or -3 but
root-9 is always 3.

I go for D. Whats the Answer?

OA is B

thanks for correcting me