Is #x^2 + 8xy - y^2# a perfect square?

1 Answer
Noah G
Mar 22, 2017

No, it isn't, because of the #-y^2#.

Explanation:

No, it isn't. A trinomial square is always of the form #a^2 +- 2ab + b^2#. In other words, parameter #b# must always be positive. This is because #n^2#, no matter the sign of #a# gives a positive value.
If you have #(-n)^2 = (-n)(-n) = n^2#.

It is in this way that #sqrt(-4) != -2#. This is because #(sqrt(-4))^2 != (-2)^2 -> -4 != 4#. I'll also use this moment to say that the #sqrt# of any negative number is undefined in the real number syste. It is in cases such as these that we talk about imaginary numbers.

Hopefully this helps!

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