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Is this statement true?

$$\sqrt{(-5)^2} = -5$$

mvw
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Nour Aldein
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  • no it is $5$ since $$(-5)^2=5^2$$ – Dr. Sonnhard Graubner Feb 18 '17 at 16:29
  • Basically square root function returns always positive value so answer should be +5.If we find sqrt of x*x it is |x| and not x. – Caratheodory_Enthusiast Feb 18 '17 at 16:31
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    We usually define $\sqrt{x}$ for $x$ real and positive to be the positive square root. But in some circtumstances, we allow for $\sqrt{x}$ to be a multi-valued function, in which $-5$ is a value for $\sqrt{(-5)^2}$. – Thomas Andrews Feb 18 '17 at 16:31
  • Also, I need to complain about your use of tags. Please read what the tags say to ensure that they mean what you intend. The tag Linear algebra is about the study of vectorspaces and linear operators between them. The tag roots is talking about roots of equations, i.e. those values of $x$ for which $f(x)=0$, for example the roots of $f(x)=(x-2)(x-3)$ are $2$ and $3$ respectively. Neither have anything to do with your question. Correct tags would have been things like Algebra-precalculus, Arithmetic, and Radicals. Do try to use correct tags in the future. – JMoravitz Feb 18 '17 at 16:40

3 Answers3

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No: The radical sign usually denotes the non-negative branch of the square root, so $$ \sqrt{x^{2}} = |x|\quad\text{ for all real $x$.} $$ Consequently, $\sqrt{(-5)^{2}} = 5$.

Andrew D. Hwang
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4

No it is not.

You have

$$\sqrt{(-5)^2}=\sqrt{(-5)\times (-5)}=\sqrt{25}=5.$$

Not $-5$.

E. Joseph
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3

$$\sqrt{(-x)^2} = |x|$$

$$(\sqrt{x})^2 = x$$

Maadhav
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