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Is this statement true?

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

  • 0
    no it is $5$ since $$(-5)^2=5^2$$ – 2017-02-18
  • 0
    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. – 2017-02-18
  • 1
    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 $\sqrt{(-5)^2}$. – 2017-02-18
  • 0
    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. – 2017-02-18

3 Answers 3

6

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

4

No it is not.

You have

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

Not $-5$.

3

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

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