Why is the square root of x^2 = |x|?

[Ebby]

Homework Statement

Why is the square root of x^2 = |x|?

Relevant Equations

sqrt(x^2) = |x|

If I reason this as follows, I run into problems. Please help me understand what is wrong with reasoning like this.

a) I start with the left hand side of the equation and let that x be -2.
b) I square it. This gives me 4. So I now have the square root of 4.
c) The square root of 4 is +/- 2. The left hand side is now +/-2.
d) I now let the x of the right hand side of the equation be -2.
e) I take the magnitude, which is 2. The right hand side is now 2.

Overall, I now have +/-2 = -2, which is wrong.

 

[haruspex]

The square root of 4 is +/- 2.

Yes, but sqrt(), or√, is a function and therefore single valued. It is defined as the positive square root.

 

[topsquark]

Homework Statement: Why is the square root of x^2 = |x|?
Relevant Equations: sqrt(x^2) = |x|

If I reason this as follows, I run into problems. Please help me understand what is wrong with reasoning like this.

a) I start with the left hand side of the equation and let that x be -2.
b) I square it. This gives me 4. So I now have the square root of 4.
c) The square root of 4 is +/- 2. The left hand side is now +/-2.
d) I now let the x of the right hand side of the equation be -2.
e) I take the magnitude, which is 2. The right hand side is now 2.

Overall, I now have +/-2 = -2, which is wrong.

It depends somewhat on what you are trying to do.

The $\sqrt{}$ function is a function: it must be single valued. By convention, we choose to take the positive branch, so $\sqrt{4} = 2$, not $\pm2$.

If you want to solve $x^2 = 4$ then what you need to do is this:
$x^2 = 4$

$x^2 - 4 = 0$

$(x + 2)(x - 2) = 0$

Thus either $x + 2 = 0 \implies x = -2$ or $x - 2 = 0 \implies x = 2$, so $x = \pm 2$.

-Dan

 

[Hill]

Homework Statement: Why is the square root of x^2 = |x|?
Relevant Equations: sqrt(x^2) = |x|

c) The square root of 4 is +/- 2.

This statement is problematic because "+/- 2" is not a number. It is a pair of numbers, namely, "+2, -2". We cannot equate a pair of numbers on one side with one number on the other side of an equation. Thus, we cannot conclude that, "The left hand side is now +/-2", and consequently, cannot conclude the rest of the reasoning, i.e., parts d) and e).

 

[jack action]

With both $x^2$ and $|x|$ you have a pair of numbers (positive and negative values) that leads you to the same answer.

On the left-hand side, you are solving $x^2$ and then reversing it, giving you the pair of numbers. To be fair, you should do the same on the right-hand side:
$$\sqrt{x^2} = ±|x|$$

 

[PeroK]

With both $x^2$ and $|x|$ you have a pair of numbers (positive and negative values) that leads you to the same answer.

On the left-hand side, you are solving $x^2$ and then reversing it, giving you the pair of numbers. To be fair, you should do the same on the right-hand side:
$$\sqrt{x^2} = ±|x|$$

This is wrong. $\sqrt x$, by definition, is a positive number.

 

[jack action]

This is wrong. $\sqrt x$, by definition, is a positive number.

Let me rewrite that for the purists while still considering the OP's questioning:
$$±\sqrt{x^2} = ±|x|$$

Every positive number $x$ has two square roots: $\sqrt{x}$ (which is positive) and $-\sqrt {x}$ (which is negative). The two roots can be written more concisely using the $±$ sign as $±\sqrt {x}$.

 

[PeroK]

Let me rewrite that for the purists while still considering the OP's questioning:
$$±\sqrt{x^2} = ±|x|$$

That says no more or less than $\sqrt{x^2} = |x|$. If $a =b$ then $-a = -b$.

 

[jack action]

That says no more or less than $\sqrt{x^2} = |x|$. If $a =b$ then $-a = -b$.

Your equation says basically $a=a$ where $a$ is the absolute value of $x$. Mine says$\{-x, x\} = \{-x, x\}$. The OP is mixing both concepts by saying $\{-x, x\} = a$.

 

[haruspex]

Your equation says basically $a=a$ where $a$ is the absolute value of $x$. Mine says$\{-x, x\} = \{-x, x\}$. The OP is mixing both concepts by saying $\{-x, x\} = a$.

Or rather, $\{-x, x\} = \{a\}$

 

[votingmachine]

For the equation x^2-1=0, there are two answers, 1 and -1.

You can solve the equation by quadratic formula:

x= (0 +/- sqrt(0-4(1)(-1)) / 2(1) =+/-sqrt(4)/2
x= +/-1

or solve by simpler factorization

(x+1)(x-1)=0

which is true for:

(-1+1)(-1-1)
and for
(1+1)(1-1)

But definitionally, the square root function is the positive answer to the question of what number squared is x^2. The quadratic formula gives the +/- outside of the square root function.

Your confusion is between the answer to what number squared is 1 and what is the square root of 1. The square root is 1. There are two numbers that when squared are 1, +1 and -1. It is a trivial difference and that is easy to get sloppy about.

 

[haruspex]

Your confusion is between the answer to what number squared is 1 and what is the square root of 1

I disagree. “The square root" is not a function. There are two square roots, just as there are n n^th roots of unity. sqrt and √ are functions, returning the positive square root.

 

[PeroK]

Your equation says basically $a=a$ where $a$ is the absolute value of $x$. Mine says$\{-x, x\} = \{-x, x\}$. The OP is mixing both concepts by saying $\{-x, x\} = a$.

That's not right, either. $\pm$ denotes an ordered pair, whereas a set is unordered.

 

[PeroK]

To summarise what I believe to be the mainstream mathematical view on this. It's important for any student to understand that for all real numbers $x$
$$\sqrt {x^2} = |x|$$Without any quibbles or embellishments.

 

[Ebby]

This was interesting to read. Thank you all for helping.