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Math Help - Square Root

  1. #1
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    Square Root

    Why is the answer to a square root problem always positive and negative?

    SAMPLE:

    sqrt{16} = -4 and +4

    Why two answers?
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  2. #2
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    Because (-4)^{2}=16 and 4^{2}=16
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  3. #3
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    Quote Originally Posted by magentarita View Post
    Why is the answer to a square root problem always positive and negative?

    SAMPLE:

    sqrt{16} = -4 and +4

    Why two answers?
    They both lead to the same answer when squared...
    (-4)^2 = 16
    4^2 = 16
    \implies \sqrt{16} = \pm 4

    Visually it may appear clearer, consider the quadratic graph of y = x^2 (Shown Below), for every \pm x value, it is mapped onto the same y value.
    Attached Thumbnails Attached Thumbnails Square Root-x-2-graph.jpg  
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    Very good reply...

    I thank both replies. I want to thank Air for the picture, which makes the answer clearer.

    By the way, does the same applies to variables?

    For exmaple:

    sqrt{x^2} = -x and +x??? True or false?
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  5. #5
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    Quote Originally Posted by magentarita View Post
    I thank both replies. I want to thank Air for the picture, which makes the answer clearer.

    By the way, does the same applies to variables?

    For exmaple:

    sqrt{x^2} = -x and +x??? True or false?
    True.

    Variables are just numbers in disguise. (A range of constants)

    \sqrt{x^2} = \left|x\right| = <br />
\begin{cases} <br />
  x,  & \mbox{if }x \ge 0 \\<br />
  -x, & \mbox{if }x \le 0 <br />
\end{cases}
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  6. #6
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    I agree that there are two square roots of any positive real number.

    But, I disagree completely that \sqrt {16}  =  \pm 4. That is simply an abuse of notation!

    This is a standard discussion in any elementary mathematics course.
    The two square roots of 16 are: \sqrt {16}  = 4\,\& \, - \sqrt {16}  =  - 4.

    Therefore, \sqrt {x^2 }  = \left| x \right|.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I agree that there are two square roots of any positive real number.

    But, I disagree completely that \sqrt {16}  =  \pm 4. That is simply an abuse of notation!

    This is a standard discussion in any elementary mathematics course.
    The two square roots of 16 are: \sqrt {16}  = 4\,\& \, - \sqrt {16}  =  - 4.

    Therefore, \sqrt {x^2 }  = \left| x \right|.
    Completely agree with that !

    If you see the graph of the function y=sqrt(x), you will see that y can't be negative.

    Actually, working with the graph y=x is a mistake because sqrt(x) is not the inverse function of x !!!!

    If one has x=16, then for sure x=+ or - sqrt(16) because x=16 --> x-16=0 --> x-(sqrt(16))=0 --> (x-sqrt(16))(x+sqrt(16))=0 and the rest follows.

    Quote Originally Posted by Air View Post
    True.

    Variables are just numbers in disguise. (A range of constants)

    \sqrt{x^2} = \left|x\right| = <br />
\begin{cases} <br />
  x,  & \mbox{if }x \ge 0 \\<br />
  -x, & \mbox{if }x \le 0 <br />
\end{cases}
    Quote Originally Posted by Air View Post
    They both lead to the same answer when squared...
    (-4)^2 = 16
    4^2 = 16
    \implies \sqrt{16} = \pm 4

    Visually it may appear clearer, consider the quadratic graph of y = x^2 (Shown Below), for every \pm x value, it is mapped onto the same y value.
    Actually, I'd say that these two messages are contradictory. You state clearly that sqrt(x)=|x|, which is true.
    So since 16=(-4)=4, sqrt(16)=|4|=|-4|=4 !
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  8. #8
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    Fabulous work!

    I thank all those who took time to help me understand this concept more and more.
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