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Can this problem be simplified further?

$\sqrt{(x+5)(x-7)}$

My initial thought was to set one side to zero and the square both sides but so I have been stuck. Its been way to long since I've had to do this and do not remember how to get rid of the radical sign.

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Mar 24 '17 at 14:26
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    $\begingroup$ As written, you do not have an equation. Note that $$(x + 5)(x - 7) = x^2 - 2x - 35 = (x^2 - 2x + 1) - 36 = (x - 1)^2 - 36$$ is not a perfect square. $\endgroup$ – N. F. Taussig Mar 24 '17 at 14:28
  • $\begingroup$ No, it can't. You can rewrite in a few other forms, which aren't simpler. $\endgroup$ – Yves Daoust Mar 24 '17 at 15:18
  • $\begingroup$ We might be able to help if you tell us the source of the question. Why do you want to "get rid of the radical sign"? Your profile suggests that this comes from a programming project. $\endgroup$ – Ethan Bolker Mar 24 '17 at 15:20
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What you are asking about is the simplification of an expression $$\sqrt{(x+5)(x-7)}$$

There is no equation to solve here, so you can not pretend that it equals zero, and then solve for $x$. That means, you need to check whether $(x+5)(x-7)$ is perhaps a square.

But, note that $$(x + 5)(x - 7) = x^2 - 2x - 35 = (x^2 - 2x + 1) - 36 = (x - 1)^2 - 36$$ and hence, $(x+5)(x-7)$ is not a perfect square of the form $y^2$, so you're stuck with the radical remaining.

Hence, $$\sqrt{(x+5)(x-7)}$$ is as simplified as it gets.

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  • $\begingroup$ Why would you even consider whether it could be a square if you can clearly see the factors. $\endgroup$ – user370967 Mar 24 '17 at 16:07
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This depends on the assumptions on the domain for $x$ and $y$. For real numbers, there is no simplification $(x+5)(x+7)=y^2$ in terms of $x$. However, should you be interested in finite fields, then over the field $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ we have $$ (x+5)(x+7)=(x+1)^2, $$ so its square root is simply $x+1$.

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  • $\begingroup$ Can you explain what you mean by assumptions and fields. $\endgroup$ – LCaraway Mar 24 '17 at 14:41
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    $\begingroup$ I think this is too advanced an idea for the OP, who's new to math.se and has tagged his question with elementary tags. $\endgroup$ – Ethan Bolker Mar 24 '17 at 15:19
  • $\begingroup$ @EthanBolker All right, I have modified the answer. $\endgroup$ – Dietrich Burde Mar 24 '17 at 15:27

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