Why this is wrong?

(-3)² = ([tex]\sqrt{9}[/tex])² --> True

-3 = [tex]\sqrt{9}[/tex] --> False

Wolfram Alpha result: False
 
Last edited:

chiro

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Because -3 doesn't equal 3?
 
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We had a thread about this recently. The point is that [tex]\sqrt{x^2}\neq x[/tex]. The square root of x is defined as the unique POSITIVE number who's square is x. Thus [tex]\sqrt{(-3)^2}[/tex] is the unique positive number who's square is [tex](-3)^2=9[/tex], obviously, this number is 3.

Thus, in general, we have that [tex]\sqrt{x^2}=|x|[/tex].

So, your first equation:

[tex](-3)^2=(\sqrt{9})^2[/tex]

is true, since [tex](-3)^2=9[/tex]and [tex]\sqrt{9}=3[/tex] and [tex]3^2=9[/tex]

But when taking the square root of both sides we get

[tex]\sqrt{(-3)^2}=\sqrt{(\sqrt{9})^2}[/tex]

which evaluates to [tex]3=\sqrt{9}[/tex], which is perfectly true!!
 

chiro

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Building on what micromass just said, use the horizontal line test to find out that there is no unique inverse except for the case when x = 0.
 
When you "undo" a squared operation, it is not as simple as just removing the little "2" superscript.

The square root of x is defined as the unique POSITIVE number who's square is x.
Really? In my classes, we've always used [tex]\sqrt[]{x^{2}}=\pm x[/tex]
 

Mentallic

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When you "undo" a squared operation, it is not as simple as just removing the little "2" superscript.



Really? In my classes, we've always used [tex]\sqrt[]{x^{2}}=\pm x[/tex]
No, this is wrong. The square root function is defined to be the unique positive square root of the number. [tex]\sqrt{9}=3[/tex] and if [tex]\sqrt{9}=\pm3[/tex] then I have no clue why we would write such answers as [tex]x^2=9 \rightarrow x=\pm \sqrt{9}[/tex] when the square root is already [tex]\pm[/tex] :wink:
 

Char. Limit

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And this is why when you use a multi-valued operator to prove something, you always have to check and see if the answer fits the original equation. Or you'll end up with extraneous or wrong solutions like -3=3.
 

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