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

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 ${a}^{2} \pm 2 a b + {b}^{2}$. 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 ${\left(- n\right)}^{2} = \left(- n\right) \left(- n\right) = {n}^{2}$.

It is in this way that $\sqrt{- 4} \ne - 2$. This is because ${\left(\sqrt{- 4}\right)}^{2} \ne {\left(- 2\right)}^{2} \to - 4 \ne 4$. 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!