# Infinite repeating problems?

1. Oct 24, 2009

### um0123

1. The problem statement, all variables and given/known data

In the problem to find the value of: http://img8.imageshack.us/img8/6827/powpic1.jpg [Broken]

You can solve it in this way:

$$x=\frac{1}{1+x}$$

where the x in 1 + x is http://img24.imageshack.us/img24/3751/powpic2.jpg [Broken]

which makes the equation:
$$x=\frac{1}{1+x}$$
$$x^2 + x = 1$$
$$x^2 + x - 1 = 0$$

2. Relevant equations
then i used quadratic equation

$$x=-\frac{1}{2} \pm \frac{\sqrt{5}}{2}$$
Use the method above to solve these equations:

1)http://img20.imageshack.us/img20/5903/powpic3.jpg [Broken]

2)http://img196.imageshack.us/img196/8639/powpic4.jpg [Broken]

3)http://img5.imageshack.us/img5/7536/powpic5.jpg [Broken]

4)MAKE UP YOUR OWN!

3. The attempt at a solution

i used quadratic equation on the problem in the setup to get:

$$x=-\frac{1}{2} \pm \frac{\sqrt{5}}{2}$$

plug that into the original equation to get

x = 0.365489 or -0.576014
===============================================

I followed this method for the first problem and got:

x=1.61803 or x=0.61803

which is kind of odd, im wondering if im doing it correctly.

Im going to proceed with the rest, but i am worried i am doing this all wrong.

Last edited by a moderator: May 4, 2017
2. Oct 24, 2009

### um0123

i did the second one but im stuck on the third, i have solved for what x is and i got:
x = sqrt(-a)
or
x= ax + sqrt(-a)

but when i plug those two into the overall equation of a/(a+x)

i get these two equations which i cant solve:

a(x^2) + ax + x(sqrt(-a)) - a = 0

and

ax + x(sqrt(-a)) - a = 0

i think i did this one wrong.

3. Oct 24, 2009

### Bohrok

$$\frac{1}{1 + \frac{1}{1 +\frac{1}{1 + \frac{1}{1 + ...}}}}$$

I'm not sure how to make it look like the right hand side here, though

You should probably leave the answers as they are without approximating them. The approximations for 1) are right, but x=0.61803 should be negative.

For the third one, did you use $$x = \frac{a}{a + x}$$ ?

4. Oct 24, 2009

### um0123

yes, for the third one i used that equation. Solved it by multiplying each side by a + x

got me x^2 + ax = a
which is x^2 +ax - a = 0

$$x = -\frac{-ax \pm \sqrt{(ax)^2 -4(1)(-a)}}{2(1)}$$

$$x = -\frac{-ax \pm \sqrt{a^2x^2 -4a}}{2}$$

$$x = -\frac{-ax \pm ax - 2\sqrt{-a}}{2}$$

THEREFORE

$$x = -\frac{-2\sqrt{-a}}{2} = -\sqrt{-a}$$

OR

$$x = -\frac{-2ax - 2\sqrt{-a}}{2} = -ax - \sqrt{-a}$$

and when i plug those two into the equation a/(a+x) i get all these weird equations.

5. Oct 24, 2009

### Bohrok

Show your steps for getting x2 + ax - a = 0. I get a different equation.

6. Oct 24, 2009

### um0123

$$x = \frac{a}{a+x}$$

multiplied each side by (a + x)

$$(a+x)x = a$$

distribute the x onto the (a+x)

$$ax+x^2 = a$$

subtract a from both sides

$$ax+x^2 - a = 0$$

rearrange

$$x^2 + ax -a = 0$$

7. Oct 24, 2009

### um0123

also i screwed up on the second one, and when i redid it i got
x = 1/2 +- sqrt(a+1)

and thats almost impossbile to plug into the equation for the second one.

8. Oct 24, 2009

### Bohrok

3) looks different from the first one. The expression looks like
$$a + \frac{a}{a + \frac{a}{a +\frac{a}{a + \frac{a}{a + ...}}}}$$
with an added a in front of the expression.

9. Oct 24, 2009

### um0123

oh, no, woops. i screwed up the drawing of the equations. its supposed to be the same as the one in the setup, except with a's instead of 1's

10. Oct 24, 2009

### Bohrok

I somehow missed your post 4. When you use the quadratic formula, you just use the constants a, b, and c in ax2 + bx + c = 0. Leave the x's out and try again.

11. Oct 24, 2009

### um0123

im sorry, i dont understand what you asked me to do.

12. Oct 24, 2009

### um0123

well, i retried 3 the equation and got two different answers.

x= -ax + sqrt(a)
or
x = sqrt (a)

but i still need to plug them into the orverall equation, and when i do i get really big and scary looking equations that i cant solve

13. Oct 24, 2009

### Bohrok

If ax2 + bx + c = 0, then
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ~not~ \frac{-bx \pm \sqrt{(bx)^2 - 4ac}}{2a}$$

The x's don't belong in the expression on the right above. Take them out in what you did and see what you get.

14. Oct 24, 2009

### um0123

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOH, wow, i completely forgot that... i feel like an idiot considering im in precalc and couldnt remember that.

anyway that gives me the answers x = a or 0, which is A LOT easier to handle than those other equations.

For problem two, i got x = 1/2 += sqrt(a+1), but im still having trouble when i plug it into
x= sqrt(a+x)
because it gives me
x = sqrt(a + 1/2 +-sqrt(a+1))
and i have square roots inside other square roots.

Last edited: Oct 24, 2009
15. Oct 24, 2009

### Bohrok

x = 1/2 += sqrt(a+1)
This isn't written correctly, and where did you get it? Why are you trying to plug it into x = √(a + x)?

16. Oct 24, 2009

### um0123

im working on problem two now whos original equation is

$$x = \sqrt{a+\sqrt{a\sqrt{a\sqrt{a+....}}}}$$

which is

$$x = \sqrt{a+x}$$

square both sides

$$x^2 = a+x$$

subtract a and x from both sides

$$x^2 - x -a = 0$$

Solve quadratically and i got

$$\frac{1}{2} \pm \sqrt{a+1}$$

17. Oct 24, 2009

### Bohrok

You forgot the 4 in the 4ac term.

It looks like you just want to verify that what you got is correct, since there's no other point in plugging one into the other. But once you correct your mistake, it's not too hard to simplify and end up with something like a=a, which I did. Since you have +-, you should take each case separately.

Last edited: Oct 24, 2009