What is the square root of x^2?

[Mark44]

How did you go from x⁴ - 16 = 0 to (x² - 4)(x² + 4) = 0?

I factored the expression on the left side.

What about the absolute value signs? I know it's a difference of square, but what if was 15 instead of 16?

I would do the same thing.

x⁴ = 15
x⁴ - 15 = 0
(x² - √15)(x² + √15) = 0
(x - $\sqrt[4]{15}$)(x + $\sqrt[4]{15}$)(x² + √15) = 0
x = $±\sqrt[4]{15}$

There are also two imaginary roots.

 

[Edgardo]

I think the confusion stems from two things:

1. Solving for x in the equation x^2 = a.
2. Using the square root function.


If you are given the equation x^2 = a, then there are two solutions, namely a positive and a negative: x = \sqrt{a} and x = -\sqrt{a}.

However, if you are using the square root function, i.e. \sqrt{x}, it only returns one value. To make this more clear, define a function

s: X \rightarrow Y
s(x) = y, such that y>0 and y^2=x.

This function s returns only one value for an input x. Replace s by the symbol \sqrt{}.

 

[tahayassen]

Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here:

{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2

If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, |-2|=2, |4|=4, |-sqrt(4)|=sqrt(4), and |sqrt(4)|=4.

 

[Mark44]

The difficulty isn't going from step 2 to step 3 - it's going from step 1 to step 2.

x² = 4
$\Leftrightarrow$ x² - 4 = 0
$\Leftrightarrow$ (x - 2)(x + 2) = 0
$\Leftrightarrow$ x = ± 2

|√4| = √4 = 2, not ±2.

BTW, tahayassen, you had essentially the same set of equations in post #28, a week ago. I missed that you had gone from
x² = 4 to
x = |√4|.

The problem here is that the first equation has two solutions, while the second equation has only one.

I have gone back and edited my post.

 

[Benn]

Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here:

{ x }^{ 2 }=4\\ x=|\sqrt { 4 } |\\ x=\pm \sqrt { 4 } \\ x=\pm 2

If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, |-2|=2, |4|=4, |-sqrt(4)|=sqrt(4), and |sqrt(4)|=4.

I think you might mean

{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2

 

[Hurkyl]

I think you might mean

{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2

As a reminder for the OP, while this argument does prove

If x² = 4, then x = 2 or x = -2​

it does not prove

x = 2 and x = -2 are both solutions to x² = 4​

unless you can argue that every step is reversible. (they are in this case)

 

[tahayassen]

As a reminder for the OP, while this argument does prove

If x² = 4, then x = 2 or x = -2​

it does not prove

x = 2 and x = -2 are both solutions to x² = 4​

unless you can argue that every step is reversible. (they are in this case)

What do you mean by every step is reversible? Can you give me an example where they are not reversible?

 

[Mark44]

x = -2 $\Rightarrow$ x² = 4

The steps here are not reversible. If x² = 4, it does not necessarily imply that x = -2.

 

[Dalek1099]

As a reminder for the OP, while this argument does prove

If x² = 4, then x = 2 or x = -2​

it does not prove

x = 2 and x = -2 are both solutions to x² = 4​

unless you can argue that every step is reversible. (they are in this case)

Wrong.Is x=2 a solution to x^2=4?-yes and is x=-2 a solution to x^2=4-yes, they are both solutions.Saying x=two values simultaneously is wrong, which I think is what you meant and thinking about this, with everything said generally being correct in this thread about(x^0.5)^2 being= to + or-(x) or abs(x) this means that (x^a)^b doesn't always equal x^(ab), which is an exponentiation law-I suppose there always are exceptions.The steps are irreversible because the inverse has more than one image, which isn't right.

 

[micromass]

Wrong.Is x=2 a solution to x^2=4?-yes and is x=-2 a solution to x^2=4-yes, they are both solutions.Saying x=two values simultaneously is wrong, which I think is what you meant and thinking about this, with everything said generally being correct in this thread about(x^0.5)^2 being= to + or-(x) or abs(x) this means that (x^a)^b doesn't always equal x^(ab), which is an exponentiation law-I suppose there always are exceptions.The steps are irreversible because the inverse has more than one image, which isn't right.

What Hurkyl said was absolutely correct. I think you misinterpreted his post.

 

[tahayassen]

Alright, thanks for the help guys.