Infinite repeating problems? Can I Solve These Quadratic Equations Correctly?

[um0123]

Homework Statement

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

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

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

Homework 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

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

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

4)MAKE UP YOUR OWN!

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, I am wondering if I am doing it correctly.

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

 

[um0123]

i did the second one but I am 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 can't solve:

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

and

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

i think i did this one wrong.

 

[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} ?

 

[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.

 

[Bohrok]

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

 

[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 sidesax+x^2 - a = 0

rearrangex^2 + ax -a = 0

 

[um0123]

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

and that's almost impossbile to plug into the equation for the second one.

 

[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.

 

[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

 

[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.

 

[um0123]

im sorry, i don't understand what you asked me to do.

 

[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 can't solve

 

[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.

 

[um0123]

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOH, wow, i completely forgot that... i feel like an idiot considering I am in precalc and couldn't 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 I am 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.

 

[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)?

 

[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}

 

[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.