Is $x^{2} + 8 x y - y^{2}$ $x^2 + 8xy - y^2$ a perfect square?

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Is $x^{2} + 8 x y - y^{2}$ $x^2 + 8xy - y^2$ a perfect square?

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Answer 1 · 2017-03-22T12:11:25

No, it isn't, because of the $- y^{2}$ $-y^2$ .

Explanation:

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

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

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Is $x^{2} + 8 x y - y^{2}$ $x^2 + 8xy - y^2$ a perfect square?

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