Recreational Problems (SM level)

[phagist_]

keep em coming please =)

(nothing too hard haha)

[Ahmad]

13. If made sense, what would equal?

14. Find the real roots of .

15. Solve .

16. Solve the equation .

[humph]

13. If made sense, what would equal?

We have .
Squaring both sides, we get

but clearly the second term is the same as the original term, so we get

[Ahmad]

Yup, here's another (though I admit less motivated, but nonetheless interesting) solution.

(n radicals).

Then take limits.

[humph]

16. Solve the equation .

Try
So 2 is a solution.

Try
So 3 is also a solution.

By inspection, this can be factorised to

So the solutions of the equation are

[Ahmad]

Right! I actually had another idea in mind, here's an extension:

17. Solve .

[humph]

Some well known but interesting ones, nevertheless:

Find the solution to

Likewise, find

More generally, find

How about   ?

[Ahmad]

This one is interesting:

18.
Where there are n radicals.

[humph]

15. Solve .

solutions are

i think.

[humph]

these can both be solved using the quadratic formula. can you solve for such that is of the form ?
(with , and .).
also can you prove that, for both answers above, is never rational?

in fact, any continued fraction that repeats itself can be solved using the quadratic formula. can you prove this?

[Ahmad]

15. Solve .

solutions are

i think.

I'm interested in a method more than solution!

in fact, any continued fraction that repeats itself can be solved using the quadratic formula. can you prove this?

If the continued fraction repeats, at some point in the continued fraction tower you will place an x instead of continuing it forever. Now to simplify the tower with x, you will perform addition and multiplication, so it simplifies into an expression of the form (ax + b)/(cx + d) which consequently results in a quadratic equation.

[humph]

15. Solve .

solutions are

i think.

I'm interested in a method more than solution!

it was basically a bit of guesswork at first  i factorised 1680, worked out what the product of the four terms must be if integers, and got the two solutions . the rest just was expanding out, shifting things around, and using the quadratic formula. certainly not a very effective method of solving the problem, it must be said.

in fact, any continued fraction that repeats itself can be solved using the quadratic formula. can you prove this?

If the continued fraction repeats, at some point in the continued fraction tower you will place an x instead of continuing it forever. Now to simplify the tower with x, you will perform addition and multiplication, so it simplifies into an expression of the form (ax + b)/(cx + d) which consequently results in a quadratic equation.

that's pretty much it. it is possible to put this into a formal proof, but it takes a lot of time and effort to write it up.

[Ahmad]

these can both be solved using the quadratic formula. can you solve for such that is of the form ?
(with a natural number, and all reals).
also can you prove that, for the first answer above, is never rational?
for the second answer above, can you prove that gives the only rational solution?

in fact, any continued fraction that repeats itself can be solved using the quadratic formula. can you prove this?

For x to be rational, then must be a square. However, for all n,
, therefore the expression is between two squares, so it cannot be a square itself!

[humph]

these can both be solved using the quadratic formula. can you solve for such that is of the form ?
(with a natural number, and all reals).
also can you prove that, for the first answer above, is never rational?
for the second answer above, can you prove that gives the only rational solution?

in fact, any continued fraction that repeats itself can be solved using the quadratic formula. can you prove this?

For x to be rational, then must be a square. However, for all n,
, therefore the expression is between two squares, so it cannot be a square itself!

precisely. same with the other one (though had to edit the question several times to get that right!)
for the second question:





again, must be a square. but

so this can never be the case.

[/0]

15. Solve .

Let





Let



or



or

Hence the solutions

edit: fixed