Kaplan DS

(1)
Square root of A^2 - B^2 is NOT equal to A-B! In fact, it has nothing to do with A-B.
As you do know yourself, (A-B)(A-B) = A^2 - 2AB + B^2, not simply A^2 - B^2

(2) A^2 - 2AB + B^2 =1
From this we can deduce either of the following:
(A-B)(A-B) = 1
or (B-A)(B-A) = 1
If the former is true, then A-B=1; if the latter is true, then A-B=-1
So statement 2 alone is not sufficient either.

Looking at the 2 statements together, you can play around a little but you will not arrive at a definite answer for A-B.

Sorry but I am still confused

I understand that (A-B)(A-B) = A^2 - 2AB + B^2 but doesn't
(A-B)(A-B) = (A-B)^2 or A^2 - B^2 if you simplifed it. Sorry im still confused. thanks

davy420 wrote:Sorry but I am still confused

I understand that (A-B)(A-B) = A^2 - 2AB + B^2 but doesn't
(A-B)(A-B) = (A-B)^2 or A^2 - B^2 if you simplifed it.

Yes
(A-B)(A-B) = (A-B)^2
but
(A-B)^2 != A^2 - B^2

rather
(A-B)^2 = A^2 + B^2 - 2AB

You can verify by multiplying (A-B) with (A-B)

Sorry im still confused. thanks

(1)a^2-b^2 = (a+b) (a-b)

a+b = 4
a-b= 9/4

a=25/8
b= 7/8

or
a + b = 3
a - b = 3

a=3 b=0

We cannot come up with a definite value for a-b INSUFFICIENT

For (2)

(A-B) (A-B) = 1

A-B CAN BE EQUAL TO -1 OR 1

INSUFFICIENT

Taking both together still INSUFFICIENT(we cannot come to a definite value
HENCE E

ohhh ok i understand now. Thanks everyone for the explaination