ilorty

Let us furst square both sides. This will give us a^2(n+1)=b^2(n-1). From here, we can subtract a^2 from both sides: na^2+a^2=nb^2-b^2 turns into na^2=nb^2-b^2-a^2.

Now, we can subtract nb^2 from both sides and get na^2-nb^2=-b^2-a^2.Now, factor out the n and divide both sides by a^2-b^2. Therefore, n in terms of b and a is: \(\frac{-b^2-a^2}{a^2-b^2}\)

Sure, I can help you, but maybe be a little more patient next time. Answerers need to take time to solve the problem.

I'm not so sure about my hint, (and I'm not too confindent in my answer) but hopefully it helps you think. For your second expression to be the smallest possible value, the values of the first equation has to have EVERY single value to be the smallest possible. Sure, you could set one value to 1/200000, but then the other values need to be larger. Using number sense, we can see that the most "logical" solution for a, b, c, and d is 1/4, 1/4, 1/4, and 1/4. Notice a pattern? That is right, they are all the same. Adding the squares up will give us (yet again!) 1/4.

oops, sniped by guest!

I checked the link, and the quesition appears to be answered. Can you check again, maybe?

Since all sides are integers, we can use pythagorean tripples to help solve this. If you dont know what that is, you can look here:

https://www.mathsisfun.com/pythagorean_triples.html (in general, mathisfun.com is a great place to search up the definition of a math term, as they give many cool diagrams and their wording is simple and easy to understand)

Anyways, we see that since BA is 3, and all sides are integers, CB has to be 4, and CA has to be 5 (3-4-5 triangle). From here, we see that since CA is the shorter leg of the blue triangle, CD has to be 12, and DA has to be 13 (5-12-13 triangle). And finally, the green triangle. Now, although DA looks like the longer leg, the only other pythagorean triple that contains 13 is a 13-84-85 triangle. So, here is a great example on why you should always use geometric intuition, but never assume. Therefore, the length of AE is 85.

If you want help, maybe you could find a coding forum for that? I only know Python, but I'm sure there are forums for questions like those, like cplusplus.com. I haven't really dug into that site yet, but it seems like it could help.

Remeber what z is! Z stands for a complex number in the form a+bi, where a and b are real, and i is an imaginary number. So, we can restate the question like:

(a+bi)^2=2i.

(a+bi)^2 can be expanded into a^2+2abi-b^2 (becase the sqaure of i is -1).

Therefore, a^2+2abi-b^2=2i. For this to be true, we see that the only way is for a^2 and b^2 to cancel out. Along with that, 2ab must equal to 2. Therefore, the options for a and b are 1 and -1. 

Using combinations, we see that the complex numbers that satisfy this equation is: 1+i, -1+i, 1-i, -1-i.

I dont think I should do your homework for you, but I can give you a hint. 

For 1)

Rotationally use the Law of Cosines until you got one of each.

Ex: First set angle B to C, then angle A, and so on.

For 2) 

Use the law of cosines (again) to get a quadratic. The larger term is a_2, and the smaller term is a_1. (Further hint, the cosine of 60 is \(\frac{\sqrt3}{2}\))

For 3)

Use the law of cosines, where 12 is c, 10 is b, and a is a. 

No problem!

If you don't understand how I found the ranges, just ask me :)