Square root simplification question: sqrt[(-2)^2 * 3]

[auryn]

I need an explanation to this square root simplification:


$\sqrt{(-2)^2*3}$

Now I know the answer is supposed to be $$2\sqrt{3}$$, but what if I split the square root into $$\sqrt{(-2)^2}\sqrt{3}$$ and then simplify just the $$\sqrt{(-2)^2}$$. I thought the square root and square were inverses, so why can't the square root cancel the square and leave -2 and thus $$-2\sqrt{3}$$ as the answer? Why does the square root cancel the square in this case $$\sqrt{x^2}=x$$ but we can't cancel it and leave a -x as the answer? I know the order of operations (PEMDAS) but why can't the square root and square cancel in this way? What am I missing? Thanks!

 

[Harry_the_cat]

Here's a question for you.

What is $\displaystyle \sqrt4$ equal to?

 

[Otis]

$\sqrt{(-2)^2}$

why can't the square root cancel the square and leave -2

Hi auryn. The radical symbol $\sqrt{}$ represents the principal square root. The principal square root is the non-negative root.

(If we desire to specify a negative root, then we need to write a negation symbol in front of the radical sign.)

$\displaystyle \text{ }\sqrt{(\text{-}2)^2} = 2$

$\displaystyle \text{-}\sqrt{(\text{-}2)^2} = \text{-}2$

Also, a radical sign is a grouping symbol, so we follow PEMDAS and simplify radicands before taking their root.

$\displaystyle \sqrt{(\text{-}2)^2} = \sqrt{4} = 2$

$\sqrt{x^2}=x$

That's not quite correct because it only works for non-negative values of x. The proper statement is:

$\displaystyle \sqrt{x^2} \;=\; |x|$


$\;$

 

[Steven G]

The reason the square root function doesn't allow negative answers is because if it did, then it would not be a function.

Sqrt(4) does equal sqrt( (-2)^2.

If y= sqrt(x) and x=4, then sqrt(4) in theory could equal both 2 and -2 as (2)^4 = 4 and (-2)^2=4. The problem is that for 1 input of x=4, we get back two output values and so sqrt(x) is not a function. To make sqrt(x) a function it was decided to reject all negative values. Now for every input value we get back at most 1 output value.

 

[auryn]

Thanks for the responses. Ok it's starting to make sense. Just to round out my understanding, the square root is limited to positive values to make it a function. Is this true for odd roots too, say the cube root?

 

[BigBeachBanana]

Thanks for the responses. Ok it's starting to make sense. Just to round out my understanding, the square root is limited to positive values to make it a function. Is this true for odd roots too, say the cube root?

What is $\sqrt[3]{-1} ~ ?$

 

[Steven G]

Thanks for the responses. Ok it's starting to make sense. Just to round out my understanding, the square root is limited to positive values to make it a function. Is this true for odd roots too, say the cube root?

Not exactly true since sqrt(0) = 0 and 0 is not positive. We say non-negative when we want 0 and positive numbers.

 

[Steven G]

Is this true for odd roots too, say the cube root?

You, not us, need to think about this. When you compute a cube root do you ever get multiple answers? If yes, then to make the cube root a function you'll need to somehow make it so that you never get more than one answer.

 

[The Highlander]

Thanks for the responses. Ok it's starting to make sense. Just to round out my understanding, the square root is limited to positive values to make it a function. Is this true for odd roots too, say the cube root?

Hi @auryn,

Until I (quite recently) joined this forum, I didn’t know about this ‘feature’ of the radical symbol (√) myself and I taught Maths (& Physics) for 30 years!

I am not a ‘mathematician” (my first degree was in Engineering) and I don’t know if you are in the UK but in my part of the world the ‘issue’ simply doesn’t arise in the High School curriculum. Our Maths teachers taught us that any non-negative number has two possible square roots (other than zero which has only the one, zero itself, of course) but it was never explained that $\displaystyle \sqrt{x}$ or, indeed, $\displaystyle x^{\frac{1}{2}}$ is defined as only the positive (ie: the Principal) square root of $\displaystyle x$.

As has been alluded to above, if these functions were not defined thus then they simply would not be functions!

I expect you are aware that the definition of a function is that for every value of $\displaystyle x$ there is a unique value for $\displaystyle f(x)$

If you graph (using, for example, Desmos or similar) $\displaystyle \sqrt{x}$ or $\displaystyle x^{\frac{1}{2}}$ and compare that to the graph of: $\displaystyle ±\sqrt{x}$ you should be able to see the ‘problem’ immediately.

The former passes the ‘vertical line test’ (a vertical line only crosses the graph at one point) whereas the latter, clearly, fails it.

So, while, $\displaystyle \sqrt{x}$ is a function,


the square root(s) of $\displaystyle x$ is not a function.
​

If you also graph the function $\displaystyle \sqrt[3]{x}$ you should then see too why this isn’t an ‘issue’ for cube roots and if you play around with various other ‘options’ it should become clear that the ‘problem’ only arises where, for $\displaystyle \sqrt[n]{x},\; n$ is an even, positive number.

NB: If you want to use Desmos then you can just type in “sqrt” or “nthroot” to graph these functions but if you want to graph an expression containing “±” then you need to include a separate line that says: “± = [1, -1]” as Desmos does not (natively) recognise the “±” symbol.

Hope that helps.

 

[Steven G]

If you graph (using, for example, Desmos or similar) $\displaystyle \sqrt{x}$ or $\displaystyle x^{\frac{1}{2}}$ and compare that to the graph of: $\displaystyle ±\sqrt{x}$ you should be able to see the ‘problem’ immediately.

The former passes the ‘vertical line test’ (a vertical line only crosses the graph at one point) whereas the latter, clearly, fails it.

@The Highlander ,
In order to sleep well tonight I feel a strong need to point out an error in your post (probably your first error on this forum). There are many vertical lines for the graph of the square root function which does NOT even cross the graph at even one point. Just look at graph that you posted.

 

[Harry_the_cat]

Thanks for the responses. Ok it's starting to make sense. Just to round out my understanding, the square root is limited to positive values to make it a function. Is this true for odd roots too, say the cube root?

No. Because cubes can be positive or negative. For example, $\displaystyle \sqrt[3]{8} = 2$ and $\displaystyle \sqrt[3]{-8} = -2$.

 

[The Highlander]

@The Highlander ,
In order to sleep well tonight I feel a strong need to point out an error in your post (probably your first error on this forum). There are many vertical lines for the graph of the square root function which does NOT even cross the graph at even one point. Just look at graph that you posted.

Not sure I understand the point you're making, Steven. ?

The idea I had hoped to get across is that if there is any vertical line that crosses a graph at more than one point then that is not the graph of a function. ?‍

But you have a good night's sleep and feel free to come back tomorrow and explain where I've gone wrong after you've properly rested; I'm always happy to take ownership of my mistakes (and there have been a few, actually*) and correct them wherever possible.??

(* Thought I'd better mention that before @Otis lists them. ?)

 

[Otis]

better mention that before @Otis lists [my mistakes]

Writing that list would take too long, so you can stop worrying.
$\;$

 

[Steven G]

Not sure I understand the point you're making, Steven. ?

The idea I had hoped to get across is that if there is any vertical line that crosses a graph at more than one point then that is not the graph of a function. ?‍

But you have a good night's sleep and feel free to come back tomorrow and explain where I've gone wrong after you've properly rested; I'm always happy to take ownership of my mistakes (and there have been a few, actually*) and correct them wherever possible.??

(* Thought I'd better mention that before @Otis lists them. ?)

When x<0, any vertical line crosses the graph in exactly 0 places.