+2441 Unfortunately, writing out such an unthinkably vast number is impossible; there are approximately \(10^{80}\) atoms in the observable universe. Googolplex is \(10^{10^{100}}=10^{10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}\). Therefore, you could write a digit on every atom and still have not scratched the surface.
Even if we assume that there are an infinite amount of atoms, we still cannot write the number out. Assuming that a normal human being can write two digits every second, it would take \(1.51*10^{91}\) years to write. That is \(1.1*10^{81}\) times the expected life expectancy of the entire universe.
The conclusion you should be gathering here is that writing out this number on a piece of paper is unfeasible.
Writing it digitally is a different story, however. Assuming a typical book is a 400 pages long, a book can print \(10^6\), or a million, zeroes on it. This would require \(10^{94}\), or 10 trigintillion, volumes before the number is written in its entirety. Then, I found this website, http://www.googolplexwrittenout.com/, that does exactly that! Write the entire number from start to finish. You can even order one of the volumes! That's crazy!
+2441 Let's see if I can simplify this expression of \(1-\frac{\sqrt{(0-PF)^2+(1-PD)^2}}{\sqrt{2}}\)
| \(1-\frac{\sqrt{(0-PF)^2+(1-PD)^2}}{\sqrt{2}}\) | Let's first notice that 0 minus a number is always itself. |
| \(1-\frac{\sqrt{(PF)^2+(1-PD)^2}}{\sqrt{2}}\) | Let's do \((PF)^2\) by using the exponent rule that \((ab)^n=a^n*b^n\) |
| \(1-\frac{\sqrt{P^2F^2+(1-PD)^2}}{\sqrt{2}}\) | Now, let's expand (1-PD)^2 by using the exponent rule that \((a-b)^2=a^2-2ab+b^2\) |
| \((1-PD)^2\) | |
| \(1^2-2(1)(PD)+(PD)^2\) | Simplify |
| \(1-2PD+P^2D^2\) | Now, reinsert this into the equation for \((1-PD)^2\). |
| \(1-\frac{\sqrt{P^2F^2+1-2PD+P^2D^2}}{\sqrt{2}}\) | Rearrange the terms in the numerator such that the terms are ordered, from left to right, based on degree. |
| \(1-\frac{\sqrt{P^2F^2+P^2D^2-2PD+1}}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}\) | Now, I want to deal with the denominator. Rationalize it by multiplying by \(\frac{\sqrt{2}}{\sqrt{2}}\). To combine the numerator and denominator, note that \(\sqrt{a}*\sqrt{b}=\sqrt{ab}\) |
| \(1-\frac{\sqrt{2(P^2F^2+P^2D^2-2PD+1)}}{2}\) | There is no common factor amongst the terms in the numerator other then 2, so there is no more simplification that can be done. |
+2441 This question has been asked an innumerable amount of times on this forum, so I will just redirect you to pages where a user has already answered this:
1. https://web2.0calc.com/questions/6-2-2-1#r2
2. https://web2.0calc.com/questions/6-2-2-1-1-or-9
3. https://web2.0calc.com/questions/6-2-1-2_5
I figure that these resources are sufficient for you!
+2441 +2441 +2441 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:
+2441 Here is another method to solve this equation. I like this method as it gets rid of all the pesky fractions and/or decimals in an equation:
| \(x=0.2x+0.6\) | Multiply both sides of the equation by 10. |
| \(10x=2x+6\) | By multiplying by 10, we have eliminated any decimals in the equation. Subtract 2x from both sides. |
| \(8x=6\) | Divide by 8 on both sides. |
| \(x=\frac{6}{8}=\frac{3}{4}=0.75\) | |
For this particular problem, the method one chooses here is trivial, as the problem is too short to make a noticeable difference. However, I find that I prefer (and have ultimately defaulted to) this method of eliminating any and all decimals and/or fractions to ease the solving process for more complicated equations.
+2441 I would argue that there is a second solution to this problem, however. Let me demonstrate why:
| \((x+a)^2=x^2+10x+25\) | Take the square root of both sides of the equation. Of course, taking the square root of a number results in a positive and a negative answer. | ||
| \(x+a=\pm\sqrt{x^2+10x+25}\) | Let's split these solutions into 2 | ||
| |||
Now, let's solve each equation separately:
| \(x+a=\sqrt{x^2+10x+25}\) | The first step is to factor the expression x^2+10x+25. Now, this trinomial is indeed a perfect-square trinomial. I know this because x^2 and 25 are perfect squares; x and 5. If you multiply the sum of x and 5 by 2, then you get 10x. If this condition is ever true with a trinomial, you have a perfect-square trinomial. |
| \(x+a=\sqrt{(x+5)^2}\) | The square root and the square cancel each other out. |
| \(x+a=x+5\) | Subtract x on both sides. |
| \(a=5\) | |
You will notice that the user above also got this answer of a=5. What about the second solution? Well, you'll see. Solve for a in the following equation \(x+a=-\sqrt{x^2+10x+25}\), the second case:
| \(x+a=-\sqrt{x^2+10x+25}\) | We have already determined previously that \(\sqrt{x^2+10x+25}\) equals \(x+5\). Let's just plug that in to save a few steps. |
| \(x+a=-(x+5)\) | Distribute the negative sign to both terms. |
| \(x+a=-x-5\) | Subtract x on both sides. |
| \(a=-2x-5\) | |
Is this actually a solution, though? If you are ever unsure of whether or not a solution is truly a solution, plug it into the original equation:
| \((x+a)^2=x^2+10x+25\) | Substitute a for -2x-5 |
| \((x+(-2x-5))^2=x^2+10x+25\) | Combine the terms of x and -2x |
| \((-x-5)^2=x^2+10x+25\) | Expand (-x-5)^2 using the rule that \((a+b)^2=a^2+2ab+b^2\) |
| \(x^2+10x+25=x^2+10x+25\) | Both sides of the equation are the same, so it should be clear now that both are equivalent. Therefore, -2x-5 is a solution. |
Therefore, there are two solutions
\(a_1=5\)
\(a_2=-2x-5\)
.