(a+b)(a-b) = a^2. b^2 hence (a+b) =a^2-b^2/a-b what if a=b:?

Answer 1

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

Hence, (a + b) = (a^2 - b^2) / (a - b)

If a = b, the above formula is invalid, because you would be dividing by 0 and would get an undefined value.

That is, a = b means a - b = 0.

Answer 2

Well if a=b, the second statement (the hence part) is proven to be false because that would make it undefined. Any value divided by 0 is undefined (if a and b are equal, their difference is zero). Thus when a=b, there is an instance in which the hence statement is false and cannot be trusted to work.

Answer 3

I hope u mean (a+b)(a-b)=a^2 - b^2.

therefore, a+b = (a^2 - b^2)/(a - b).

since when a=b the LHS becomes 0/0 form, we can't use directly, but instead we can apply the limit on both sides that a tends to b. that is a is very very close to b but not equal to b.
hence as someone stated before, we can LHS as (a+b)(a-b)/(a-b) and hence = a+b.

Answer 4

if a=b then the numerator of lhs becomes a^2-b^2=o
and the denominator a-b=o.
therefore the lhs becomes =0/0 which is an invalid mathematical term. hence for a=b the eqution is not valid

Answer 5

(a+b)(a-b) = a^2. b^2 hence (a+b) =a^2-b^2/a-b what if a=b:?
I assume you mean (a+b)(a-b) = a^2 - b^2.
Hence a+b = (a^2 -b^2)/(a-b)
So a+b = [(a+b)(a-b)]/(a-b) = a+b

By canceling out the denominator we avoid dividing by 0 when a=b.

Answer 6

If a = b, then a - b = 0. Thus, your second equation would be undefined because it would have a 0 in the denominator, which isn't allowed.

Answer 7

in u r statement :
(a+b)(a-b) = a^2. b^2 is not correct i think (a+b)(a-b) = a^2 - b^2.

if it is true then if a=b then the answer is not defined since 0/0=undefined.

Answer 8

When a=b, then
2a*2b=2a*2a=4a^2
But a^2-b^2=a^2-a^2=0
Therefore the id will not be valid.
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Answer 9

(a+b)(a-b)=a^2-b^2.....for any values of a&b
But,(a+b)=a^2-b^2/(a-b)....for any a not equal to b

Answer 10

answer will be zero