Taniuca wrote:I really need help with exponents and roots.
Could someone please explain how for 1) a=b or -a=b I need some good help!! thanks!

What is the value of a?
1) a^2=b^2
2) b-a=-12

a=?

1) a^2=b^2

This one is insufficient because a,b could be negatives or positives or a single number.

-2^2 = 2^2 or 2^2 = 2^2

2) b-a=-12

This one is also insufficient.
-6 - (+6) = -12
-3 - (+9) = -12
or other numbers.


Together: They are sufficient. We know from statement 1 that |a=b|. Therefore a^2 = b^2

Since a=b, from statement 2 we know it has to be -6 - (+6) = -12.

Gurpinder,
how do you assumed that |a=b|?
thanks!

Taniuca wrote:Gurpinder,
how do you assumed that |a=b|?
thanks!

a^2=b^2 the only way this is true is if A and B are the same. otherwise this wont work. But A,B can be negatives or positives because when a number is squared, it turns into a positive.

-2^2=4
2^2=4

if A = -2 and B = 2....... a^2=b^2 ---> 4=4

|A=B| < -- i am taking their absolute value regardless of the sign.

Does that make sense?

Taniuca wrote:I really need help with exponents and roots.
Could someone please explain how for 1) a=b or -a=b I need some good help!! thanks!

What is the value of a?
1) a^2=b^2
2) b-a=-12

From Stmt1: since a^2=b^2.We can conclude that a=b or a=-b.Since a negative number raised to a even power always yields a positive number.So,it does not matter whether a,b are negative or positive.But stmt1 alone is insufficient to get a value of a.

Stmt2:b-a = -12; for this to hold true b,a can be (-9,3);(-6,6);and so on.Stmt 2 is not sufficient to yield a unique value for a.
Combining I and II we get that the only possible values for b,a=(-6,6),as I shall also hold good.
Thus IMO:answer is C