MitPY 10: Square Roots

December 2, 2020

8:19 pm

indices, Mathematical Methods, MitPY, roots, Specialist Mathematics, tests

This MitPY comes from a student, Jay:

I have a question relating to polynomial equations. For context I have just finished Y11 during which I completed Further 3&4, Methods 1&2 and Specialist 1&2.

This year during my maths methods class we covered the square root graph, however I was confused as to why it only showed the positive solutions. When I asked about it I was told it was because the radical symbol meant only the positive solution.

However since then I have learnt that the graph of $\boldsymbol{y=x^{0.5}}$ also only shows the positive solution of the square root, while $\boldsymbol{y^2=x}$ shows both. I am quite confused by why they aren’t the same. The only reason that I could think of is that it would mean $\boldsymbol{y=x^2}$ would be the same as $\boldsymbol{y^2=x^4}$ , and while the points (-2,-4) and (2,-4) fit the latter they clearly don’t fit former.

Could you please explain why these aren’t the same?

18 Replies to “MitPY 10: Square Roots”

Red Five says:

December 2, 2020 at 8:36 pm

I’ll start with a brief comment and then leave it to others, more knowledgeable…

If $x=y^{2}$ and $y \geq 0$ then the $x=y^{2} \Leftrightarrow y=\sqrt{x}$

However if $x=y^{2}$ and $y \in R$ then $x=y^{2} \Leftarrow y=\sqrt{x}$ and not the other way.

Reply
V says:

December 2, 2020 at 8:44 pm

Basically the exponential function ( e^x ) is defined as the sole curve satisfying the requirements that:
(a) its it’s own derivative.
(b) it is 1 when x = 0.
This curve only has one branch. (The positive branch.)

As x^a = e^(ln(x)*a), hence the range of x^a is R+.
(The natural logarithm function can be defined as the inverse of the exponential function, and hence will only gives a single value.)

Reply
John Friend says:

December 2, 2020 at 8:56 pm

The meaning attached to $\sqrt{x}$ where $x > 0$ is the positive number that when squared is equal to $x$ . That is, in the real context $\sqrt{x}$ is used to distinguish the positive root. So $\sqrt{4} = 2$ , not $\pm 2$ .

This is different to $x^2 = 4 \implies x = \pm \sqrt{4} = \pm 2$ . So $y^2 = x \implies y = \pm \sqrt{x}$ .

Reply
1. Jay says:
  
  December 3, 2020 at 5:03 pm
  
  Hi John,
  Thanks for you answer,
  My problem is that while I understand why $\sqrt x$ only gives the postive solution I don’t understand why $x^{0.5}/x^{1/2}$ does not.
  
  Reply
  1. John Friend says:
    
    December 3, 2020 at 6:02 pm
    
    Are you drawing these graphs using software? Plotting software interprets $\displaystyle \sqrt{x}$ and $\displaystyle x^{1/2}$ etc. as the same things. But they are not. $\sqrt{x}$ is used to distinguish the positive root, whereas $\displaystyle x^{1/2}$ is a multi-valued function. $\sqrt{x}$ is the principal valued root. See p22 of attachment.
    
    Brown-Churchill Complex Variables
    
    Reply
    1. Jay says:
      
      December 8, 2020 at 3:32 pm
      
      For graphing I usually use either my CAS or desmos, although I have had problems with them not being able to do what I what them to do before.
      Would you have any suggestion of software that doesn’t have / has less limitations like this?
      
      Reply
      1. John (No) Friend (of VCAA) says:
        
        December 8, 2020 at 6:30 pm
        
        1. You’ll be using the CAS-calculator in your Units 3&4 maths assessment and exams, so pragmatically it’s best to keep using it, be familiar with it and in particular be aware of its limitations.
        2. Software that won’t have as many limitations will be expensive due to licensing fees, and it’s not what you’ll be using in your exams.
        
        Reply
tom says:

December 2, 2020 at 9:09 pm

Dear Jay
It is perhaps simpler than you feared. Mathematicians have agreed that the square root sign just signifies the positive value. No great reason, just that it is sometimes handy to mean a specific value. So $\sqrt{4}$ is $2$ , not $\pm 2$ . [Best to use a root sign here rather than a 1/2 power.]

If we solve an equation
$x= y^2$ for $y$ , the general solution is $y=\pm \sqrt{x}$ ; the $\pm$ is needed because the $\sqrt{x}$ would only give positive $y$ (or 0). So although the graph of $x=y^2$ is a full parabola tipped on its side, the graph of $y=\sqrt{x}$ is just the top part of the parabola.

Now to complicate matters, if and when you learn complex function theory, the notation $x^{1/2$ might be used to denote both answers – so if $f(x) = x^{1/2}$ then $f(4)=\pm 2$ . Here we use just the index notation, not the root sign. By this stage mathematics has escape from the silly convention that a function can only take 1 value.

Reply
1. Jay says:
  
  December 3, 2020 at 4:37 pm
  
  Hi Tom,
  Thanks for the reply,
  So I get why $\sqrt x$ only gives the postivie solution. But what I’m confused by is why $x^{1/2}$ / $x^{0.5}$ also only gives the positive solution. However are you saying that it does give both solutions? And if so would this mean that y^2=x^4 and y=x^2 should also be the same?
  
  Reply
  1. John Friend says:
    
    December 3, 2020 at 6:07 pm
    
    $\displaystyle y^2 = x^4 \implies y = \pm x^2$ .
    
    Note Glen’s comment below. In particular:
    
    “y = x^2 implies y^2 = x^4. This implication does *not* go the other way”.
    
    In other words y^2 = x^4 does NOT imply y = x^2.
    
    Reply
Glen says:

December 3, 2020 at 8:45 am

The other comments already cover it mostly, I’ll just add a remark related to your latter comment. When we are talking about an equality such as

$y = x^2$

where $x$ and $y$ are real numbers (we write $``x,y\in\mathbb{R}''$ ), think for a moment what it means to represent that visually. What we are doing is putting a dot down at each point $(x,y)$ that satisfies this relationship. That gives us a collection of points $(x,y)$ , all of which satisfy the equality $y = x^2$ .

This is the general procedure to finding all the solutions to an equation (equality or otherwise) and doesn’t rely on the equation being linear, or even being “solved” for $y$ . An example is your second one:

$y^2 = x^4\,.$

We can use the same philosophy and look for the collection of points $(x,y)$ such that $y^2 = x^4$ ; for each one, we plot a point. If we do this carefully, we will find that the solution set for the second equality *contains* the solution set for the first equality. This is because logically $y = x^2$ implies $y^2 = x^4$ . This implication does *not* go the other way, however, and so what we find is that the solution set for the second equality is strictly larger than the first. Since the manipulation that goes from one to the other is quite straightforward, we can write the solution set for the second equality in terms of copies of the first.

I have avoided the use of set notation (set builder notation is quite handy in this setting) here, but if you are familiar with that, it does allow for a rather neat looking expression for this last point.

Reply
1. Jay says:
  
  December 3, 2020 at 4:58 pm
  
  Hi Glen,
  Thanks for your answer,
  When manipulating 2 sides of an equation as long as both side are manipulated in the same way, I thought the possible results should stay the same, are you saying that is incorrect?
  
  Reply
  1. John Friend says:
    
    December 3, 2020 at 6:12 pm
    
    Re: “When manipulating 2 sides of an equation as long as both side are manipulated in the same way, I thought the possible results should stay the same, are you saying that is incorrect?”
    
    Yes. Consider the simple equation x = 1. Now ‘manipulate’ both sides the same way by squaring: x^2 = 1.
    
    x^2 = 1 does not give back x = 1, it gives $x = \pm 1$ .
    
    Reply
Storyteller says:

December 3, 2020 at 8:21 pm

Hello Jay,

Consider this as Part 1.
You might already have a good handle on this, but I will post it anyway given I have written it.

Here is how I think about the number, $\sqrt{x}$ .

Firstly $\sqrt{9}=3$ . Not +3 and -3. Which makes some sense immediately, because if $\sqrt{x}$ is a number, it cannot have more than one value.

But why is this so?

Early on we hopefully learn that if you square +3 you get the same result as you do if you square -3. So we can appreciate that all positive real numbers have two $\textbf{square roots}$ , one positive and one negative.
E.g. 9 has 2 square roots, +3 and –3; 3 x 3 = 9 and -3 x -3 = 9.

What about 17? It also has two square roots, a positive number, a bit more than 4, and, its negative friend, – (a bit less than 4).

In general, a positive integer $k$ has two square roots, $\emph{a positive number that when squared gives k}$ , and its negative friend.

So some values of $k$ have integer square roots and some do not.
How should we write down the non-integer ones? Decimals? Nooooo!)

One way would be to create a notation, for $\emph{the positive square root of k}.$ , say $\sqrt{k}$ and its negative friend would then be $-\sqrt{k}$

$\sqrt{k}$ also got given a name other than the $\emph{the positive square root of k}.$ It was dubbed the $\emph{principal square root of k}.$

End of Part 1.

Reply
1. Storyteller says:
  
  December 3, 2020 at 8:26 pm
  
  Start of Part 2.
  
  Jay, can you say why you think they are the same?
  
  Reply
  1. Storyteller says:
    
    December 4, 2020 at 6:03 am
    
    Part 2 cont.
    
    Think about the equation $y^2=x$ like this:
    it will be true when $y$ is a number that when squared results in $x$ and there are two such numbers, the two square roots of $x$ .
    Using the notation from Part 1, $+\sqrt{x}$ and $-\sqrt{x}$ .
    
    So we say that solving $y^2=x$ for $y$ (as others have said) gives $y=\pm\sqrt{x}$ .
    In solving this equation we have not $\sqrt{}$ -ed both sides. Rather, thought about what $y$ can be.
    
    Where $y^2=x^4$ is concerned, if solving for $y$ , we want to be numbers that when squared give $x^4$ , so the two square roots of $x^4$ , so $+\sqrt{x^4}$ and $-\sqrt{x^4}$ or $+x^2$ and $-x^2$ .
    
    So, as others have said, $y^2=x^4$ $\Rightarrow {y=}\pm$ $x^2$ .
    
    Does that help?
    
    Reply
    1. John Friend says:
      
      December 4, 2020 at 6:32 am
      
      An alternative way of thinking about $y^2 = x^4$ :
      
      $y^2 = x^4 \implies y^2 - x^4 = 0 \implies (y - x^2)(y + x^2) = 0$ .
      
      From the null factor law either
      $y - x^2 = 0 \implies y = x^2$
      *or*
      $y + x^2 = 0 \implies y = -x^2$ .
      
      An alternative way of thinking about $y^2 = x$ :
      
      $y^2 = x \implies y^2 - x = 0 \implies (y - \sqrt{x})(y + \sqrt{x}) = 0$ .
      
      From the null factor law either
      $y - \sqrt{x} = 0 \implies y = \sqrt{x}$
      *or*
      $y + \sqrt{x} = 0 \implies y = -\sqrt{x}$ .
      
      Reply
      1. Jay says:
        
        December 8, 2020 at 3:25 pm
        
        Hi Storyteller and John,
        
        Thanks for your help, the explanations have been really helpful and I understand it now.
        
        Reply

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