- #1

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[tex]x = -5[/tex]

[tex]x^2 = (-5)^2[/tex]

[tex]x^2 = 25[/tex]

[tex]x = 5[/tex]

So in squaring the negative I remove it, and then when I root the number back to original magnitude, the negative has disappeared. What am I missing?

- Thread starter mpatryluk
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- #1

- 46

- 0

[tex]x = -5[/tex]

[tex]x^2 = (-5)^2[/tex]

[tex]x^2 = 25[/tex]

[tex]x = 5[/tex]

So in squaring the negative I remove it, and then when I root the number back to original magnitude, the negative has disappeared. What am I missing?

- #2

- 146

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(-1)(-1)=1=(1)(1)

- #3

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If you are presented with ## x^2 = 25 ## and asked to take the square root, your answer would be ## x = \pm 5 ##.

[tex]x = -5[/tex]

[tex]x^2 = (-5)^2[/tex]

[tex]x^2 = 25[/tex]

[tex]x = 5[/tex]

So in squaring the negative I remove it, and then when I root the number back to original magnitude, the negative has disappeared. What am I missing?

If you are in some situation where you know the answer has to be positive or negative you would have to offer some clarification in writing (or explanation if you are demonstrating to someone). Sometimes, for example, in physical problems, you know you have a positive magnitude, so the negative answer might not have any physical meaning.

-Dave K

- #4

HallsofIvy

Science Advisor

Homework Helper

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Strictly speaking, no. If you were asked toIf you are presented with ## x^2 = 25 ## and asked to take the square root, your answer would be ## x = \pm 5 ##.

The real difficulty is that the "squaring function", [itex]f(x)= x^2[/itex], is NOT "one-to-one" and so does not have a true "inverse". In particular, while it is always true that [itex](\sqrt{x})^2= x[/itex], it is NOT always true that [itex]\sqrt{x^2}= x[/itex]. What is true is that [itex]\sqrt{x^2}= |x|[/itex].

If you are in some situation where you know the answer has to be positive or negative you would have to offer some clarification in writing (or explanation if you are demonstrating to someone). Sometimes, for example, in physical problems, you know you have a positive magnitude, so the negative answer might not have any physical meaning.

-Dave K

- #5

Mark44

Mentor

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$$ x = 5$$

$$ \Rightarrow x^2 = 5^2 = 25$$

$$\Leftrightarrow x = \pm 5$$

The squaring operation in the second equation is not one-to-one, and so not reversible, which is the reason for the "implies" one-way arrow. Another way to think of this is that the solution set of the first equation {5} is different from the solution set of the second equation, which is {5, -5}. The solution sets of the second and third equations are exactly the same, which is indication by the double-ended arrow.

The operations that are one-to-one, and hence reversible, are the arithmetic operations (add, subtract, multiply, divide - restricted to nonzero divisors), exponentiating (taking e to some power), taking the log, and so on.

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