#1**+1 **

You are wondering how to graph the following equation of \(y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\). First, we should determine the domain to see what the real inputs are:

\(y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Factor out a 2 from 2x+2 |

\(y=\sqrt{2(x+1)\sqrt{x^2-1}}-\sqrt{x-1}\) | Also, x^2-1 can be factored into (x+1)(x-1). |

\(y=\sqrt{2(x+1)\sqrt{(x+1)(x-1)}}-\sqrt{x-1}\) | Split the 2 out of the radical. |

\(y=\sqrt{2}\sqrt{x+\sqrt{(x+1)(x-1)}}-\sqrt{x-1}\) | |

Let's think about which values for *x* will result in a nonreal answer. The square root of a negative number results in a nonreal answer. Let's figure out when the radicand is less than 0. Let's tackle the easy one first.

\(x-1<0\) | Add 1 to both sides of the equation. |

\(x<1\) | |

Therefore, we know that when x<1, it is not apart of the domain. Now, let's tackle the hard one:

\((x+1)(x-1)<0\) | Let's calculate when both factors are less than 0 and then calculate solutions. | ||

| Add or subtract 1 from both sides. | ||

| |||

I have not solved the inequality completely, but we know that the solutions must be something less than 1. However, we have already determined that a number less than 0 would result in a non-real output, so there is no reason to consider this inequality:

\(\text{Domain:}\{\hspace{1mm}f(x)\in\mathbb{R}|x\geq1\}\)

Now that we have determined the domain of the function, let's begin plotting some points. By determining the domain, we are eliminating the possibility of plugging in a number not in the domain. However, before we begin plotting, we should be familiar with the parent function of \(f(x)=\sqrt{x}\). I have a picture below that illustrates it:

Let's plug in the first available value into the equation, 1. That is part of the domain, so let's plug it in and see what we get:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Replace every instance of x with one. |

\(f(1)=\sqrt{2*1+2\sqrt{1^2-1}}-\sqrt{1-1}\) | Simplify. |

\(f(1)=\sqrt{2+2\sqrt{0}}-\sqrt{0}\) | Of course, \(\sqrt{0}=0\) |

\(f(1)=\sqrt{2}\) | |

Therefore, one point on the graph is \((1,\sqrt{2})\). Let's find another. Let's plug in 2:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Replace every instance of x with two. |

\(f(2)=\sqrt{2*2+2\sqrt{2^2-1}}-\sqrt{2-1}\) | Simplify |

\(f(2)=\sqrt{4+2\sqrt{3}}-\sqrt{1}\) | \(\sqrt{1}=1\), of course. Now, let's simplify this monstrosity\(\sqrt{4+2\sqrt{3}}\) |

\(\sqrt{4+2\sqrt{3}+\sqrt{3}^2-3}\) | Maybe you can that I am not changing the value of this expression by adding those 2 terms. The significance of which will become clear soon. Simplify. |

\(\sqrt{\sqrt{3}^2+2\sqrt{3}+1}\) | It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. |

\(\sqrt{(\sqrt{3}+1)^2}\) | The square root and the square function undo each other. |

\(\sqrt{3}+1\) | Now, let's replace \(\sqrt{4+2\sqrt{3}}\) with its simplified value. |

\(f(2)=\sqrt{3}+1-1\) | |

\(f(2)=\sqrt{3}\) | |

Therefore, another point is \((2,\sqrt{3})\). Let's one more example. I know ahead of time that x=5 is a good point to plug in:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Just like before, substitute x for 5. |

\(f(5)=\sqrt{2*5+2\sqrt{5^2-1}}-\sqrt{5-1}\) | Simplify. |

\(f(5)=\sqrt{10+2\sqrt{24}}-2\) | Let's try to do the same process like above to simplify the radical: |

\(\sqrt{10+2\sqrt{24}}\) | |

\(\sqrt{10+2\sqrt{24}}\) | Rearrange the terms to put them in the form of a^2+2ab+b^2. Before we do that, however, we must simplify the square root of 24. |

\(\sqrt{10+4\sqrt{6}}\) | Now, do the same process as above |

\(\sqrt{10+4\sqrt{6}+\sqrt{6}^2-6}\) | Rearrange the terms and simplify. |

\(\sqrt{\sqrt{6}^2+4\sqrt{6}+4}\) | Yet again, this is a perfect square trinomial. Factor it. |

\(\sqrt{(\sqrt{6+2})^2}\) | Just like above, the square root and the square functions cancel. |

\(\sqrt{6}+2\) | Plug this in. |

\(f(5)=\sqrt{6}+2-2\) | |

\(f(5)=\sqrt{6}\) | |

Here's a third point \((5,\sqrt{6})\). Not every integer will be nice solutions. These, however, are examples of the nicest examples that you will get. After plotting these 3 points, connect them loosely like the shape of the parent function. I picked those points because I knew that they all would have a nice answer. Other points you plug in may not have as nice of a solution.

If you would like to reference a graph, I have one for you! Click here to view it:

TheXSquaredFactor
Aug 7, 2017

#1**+1 **

Best Answer

You are wondering how to graph the following equation of \(y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\). First, we should determine the domain to see what the real inputs are:

\(y=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Factor out a 2 from 2x+2 |

\(y=\sqrt{2(x+1)\sqrt{x^2-1}}-\sqrt{x-1}\) | Also, x^2-1 can be factored into (x+1)(x-1). |

\(y=\sqrt{2(x+1)\sqrt{(x+1)(x-1)}}-\sqrt{x-1}\) | Split the 2 out of the radical. |

\(y=\sqrt{2}\sqrt{x+\sqrt{(x+1)(x-1)}}-\sqrt{x-1}\) | |

Let's think about which values for *x* will result in a nonreal answer. The square root of a negative number results in a nonreal answer. Let's figure out when the radicand is less than 0. Let's tackle the easy one first.

\(x-1<0\) | Add 1 to both sides of the equation. |

\(x<1\) | |

Therefore, we know that when x<1, it is not apart of the domain. Now, let's tackle the hard one:

\((x+1)(x-1)<0\) | Let's calculate when both factors are less than 0 and then calculate solutions. | ||

| Add or subtract 1 from both sides. | ||

| |||

I have not solved the inequality completely, but we know that the solutions must be something less than 1. However, we have already determined that a number less than 0 would result in a non-real output, so there is no reason to consider this inequality:

\(\text{Domain:}\{\hspace{1mm}f(x)\in\mathbb{R}|x\geq1\}\)

Now that we have determined the domain of the function, let's begin plotting some points. By determining the domain, we are eliminating the possibility of plugging in a number not in the domain. However, before we begin plotting, we should be familiar with the parent function of \(f(x)=\sqrt{x}\). I have a picture below that illustrates it:

Let's plug in the first available value into the equation, 1. That is part of the domain, so let's plug it in and see what we get:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Replace every instance of x with one. |

\(f(1)=\sqrt{2*1+2\sqrt{1^2-1}}-\sqrt{1-1}\) | Simplify. |

\(f(1)=\sqrt{2+2\sqrt{0}}-\sqrt{0}\) | Of course, \(\sqrt{0}=0\) |

\(f(1)=\sqrt{2}\) | |

Therefore, one point on the graph is \((1,\sqrt{2})\). Let's find another. Let's plug in 2:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Replace every instance of x with two. |

\(f(2)=\sqrt{2*2+2\sqrt{2^2-1}}-\sqrt{2-1}\) | Simplify |

\(f(2)=\sqrt{4+2\sqrt{3}}-\sqrt{1}\) | \(\sqrt{1}=1\), of course. Now, let's simplify this monstrosity\(\sqrt{4+2\sqrt{3}}\) |

\(\sqrt{4+2\sqrt{3}+\sqrt{3}^2-3}\) | Maybe you can that I am not changing the value of this expression by adding those 2 terms. The significance of which will become clear soon. Simplify. |

\(\sqrt{\sqrt{3}^2+2\sqrt{3}+1}\) | It might be difficult to notice this, but the radicand is actually a perfect-square trinomial. |

\(\sqrt{(\sqrt{3}+1)^2}\) | The square root and the square function undo each other. |

\(\sqrt{3}+1\) | Now, let's replace \(\sqrt{4+2\sqrt{3}}\) with its simplified value. |

\(f(2)=\sqrt{3}+1-1\) | |

\(f(2)=\sqrt{3}\) | |

Therefore, another point is \((2,\sqrt{3})\). Let's one more example. I know ahead of time that x=5 is a good point to plug in:

\(f(x)=\sqrt{2x+2\sqrt{x^2-1}}-\sqrt{x-1}\) | Just like before, substitute x for 5. |

\(f(5)=\sqrt{2*5+2\sqrt{5^2-1}}-\sqrt{5-1}\) | Simplify. |

\(f(5)=\sqrt{10+2\sqrt{24}}-2\) | Let's try to do the same process like above to simplify the radical: |

\(\sqrt{10+2\sqrt{24}}\) | |

\(\sqrt{10+2\sqrt{24}}\) | Rearrange the terms to put them in the form of a^2+2ab+b^2. Before we do that, however, we must simplify the square root of 24. |

\(\sqrt{10+4\sqrt{6}}\) | Now, do the same process as above |

\(\sqrt{10+4\sqrt{6}+\sqrt{6}^2-6}\) | Rearrange the terms and simplify. |

\(\sqrt{\sqrt{6}^2+4\sqrt{6}+4}\) | Yet again, this is a perfect square trinomial. Factor it. |

\(\sqrt{(\sqrt{6+2})^2}\) | Just like above, the square root and the square functions cancel. |

\(\sqrt{6}+2\) | Plug this in. |

\(f(5)=\sqrt{6}+2-2\) | |

\(f(5)=\sqrt{6}\) | |

Here's a third point \((5,\sqrt{6})\). Not every integer will be nice solutions. These, however, are examples of the nicest examples that you will get. After plotting these 3 points, connect them loosely like the shape of the parent function. I picked those points because I knew that they all would have a nice answer. Other points you plug in may not have as nice of a solution.

If you would like to reference a graph, I have one for you! Click here to view it:

TheXSquaredFactor
Aug 7, 2017

#2**0 **

Very crafty, X^{2} ....that one took a bit of work !!!!

I particularly like that trick of producing a perfect square from the sum of an integer and a radical multiplied by an integer !!!.....where did you learn that???

CPhill
Aug 7, 2017

#3**0 **

The trick does not always work, unfortunately. I just got very lucky that all of those examples actually simplify to something nice. Although I cannot locate where exactly I learned this awesome trick, I have located something that alters the method slightly, but it should give you an idea of what I am doing:

TheXSquaredFactor
Aug 7, 2017

#4**0 **

When I learned this trick, I was stunned, too. This cannot be done in every circumstance, however. I just got lucky that it happened to worked on every occasion. Although i cannot locate where I originally learned this method, I can provide you to a link where someone utilizes this method to simplify a complex radical expression. Here it is:

TheXSquaredFactor
Aug 8, 2017