Help with an exercise

[latent snow]

x=(shows in the picture); I did a) and x = the square root of 2. but I think it needs to be equal to - the square root of 2

So that you can simplify it

because I'm pretty sure it needs to end up (-1) to the power of 2023

Can someone point out my mistake?

[latent snow]

I also apologize for the language in this picture.

a) Solve square x

[manic stratus]

Note the following:
√(3 ± √(5)) = (1/√(2))√(6 ± 2√(5)) = (1/√(2))√(5 ± 2√(5) + 1) = (1/√(2))√((√(5) ± 1)^2) = (√(5) ± 1)/√(2)

[latent snow]

is there any way to illustrate this equation? I am pretty confused by looking at it.

$$√(3 ± √(5)) = (1/√(2))√(6 ± 2√(5)) = (1/√(2))√(5 ± 2√(5) + 1) = (1/√(2))√((√(5) ± 1)^2) = (√(5) ± 1)/√(2)$$

[austere ore]

mircea-

[manic stratus]

Basically, what I'm doing is noticing that the expression inside the square root can become a square, since:
6 ± 2√(5) = 5 ± 2√(5) + 1 = (√(5))^2 ± 2*√(5)*1 + 1^2 = (√(5) ± 1)^2

[latent snow]

I see, but how does this help me solve for x squared?

[manic stratus]

Well, if √(3 + √(5)) = (√(5) + 1)/√(2) and √(3 - √(5)) = (√(5) - 1)/√(2), then you can immediately get a simple value for x.

After all, there are no nested square roots now.

[latent snow]

Oh, okay, thank you

by the way

[faint meadow]

@latent snow has given 1 rep to @manic stratus

[latent snow]

one question

after solving

x = ±√2

I can use two versions for (b)

The logical one is -√2 because it simplifies

But because there are 2 solutions

I actually should solve for both, shouldn't I?

@manic stratus

[manic stratus]

No.

Look carefully at the definition of x.

Which number is greater, √(3 - √(5)) or √(3 + √(5))?

[latent snow]

Ohhh, rightt, thank you so much

[faint meadow]

@latent snow has given 1 rep to @manic stratus

[manic stratus]

You're welcome!

[latent snow]

Could you help me with one more exercise? Instead of me creating another post

"Decompose the expression into the product of three factors"

[manic stratus]

Oh, interesting! Well, let's try...
First thing is the glaring sum of squares. Let's apply that, and also factor the first two terms.
a^2 b + ab^2 + (2ab + a^2 + b^2) + a + b = ab(a + b) + (a + b)^2 + (a + b)
Now we can factor out (a + b).
ab(a + b) + (a + b)^2 + (a + b) = (a + b)(ab + a + b + 1)
Then we notice that (ab + a) + (b + 1) = a(b + 1) + (b + 1) = (a + 1)(b + 1). So:
(a + b)(ab + a + b + 1) = (a + 1)(b + 1)(a + b)
So, E = (a + 1)(b + 1)(a + b).

[latent snow]

Thank you!!!

[faint meadow]

@latent snow has given 1 rep to @manic stratus

[latent snow]

Took me a few minutes as you can see..

[manic stratus]

Glad you got it! How did you do it?

[latent snow]

Just like you but I had to get some ideas from you, I need to solve more exercises like these because it was obvious that there was a square formula, I guess I'm just for now pretty bad at spotting things like that but I'm sure I'll get better in time

Sorry for the late reply also

My math teacher gave me a book where there are problems from my country from 1957-2001

I tried solving one

Turned my ego back at 0

[latent snow]

olympiad*

[manic stratus]

Well, that's why they are more difficult 😄
Well, practice should help. Though, I am not well versed in olympiad problems, so not sure what to read for that.

[latent snow]

I also use quora sometimes for more answers when I don't quite understand from somewhere else the explanation, want me to show you the problem I was talking about by the way?

[manic stratus]

Sure!

[latent snow]

If these numbers are distinct and form a arithmetic progression, demonstrate that b = d.

[manic stratus]

I think you forgot the picture or part of the statement.

[latent snow]

Yeah yeah

[manic stratus]

Ah, ok.

[latent snow]

I took it on my phone

[manic stratus]

Hm, looks interesting! Let's see.
First, as this must be an arithmetic progression, we must have a constant difference of terms. So:
(a + c)/(b + d) - a/b = c/d - (a + c)/(b + d)
a/b + c/d = 2(a + c)/(b + d)
Let's multiply everything by (b + d).
ad + bc = 2(a + c)
But then I don't know what else to do.