fraction within fraction algebra: sqrt(5)+2 = 4 + 1/(4 + 1/(4 + 1/(sqrt(5)-a))); find "a"

[nanase]

I am currently working on this basic algebra/arithmetic problem

I tried simplifying from the last but ended up getting something I can't simplify further

the answer said it should be -2.
Can somebody tell me if I made an error in the step, or is there a better way for solving this?
Thanks ^^

 

[Dr.Peterson]

It would be a very good idea to distribute that 4:

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Alternatively, you could solve without simplifying, just repeatedly subtracting 4 and taking reciprocals (and rationalizing denominators). That will work very nicely.

 

[nanase]

Thank you Dr Peterson, I tried your interesting way of subtracting by 4, taking reciprocal and multiplying with conjugates. I got my last equation as [math]\sqrt{5} + 2 = \sqrt{5} - a[/math]which gives a as -2.
Very enlightened to see the process and getting the answer, credit to you. Thanks!

 

[BigBeachBanana]

Notice the RHS can be written as
[math]x = 4 + \dfrac{1}{x} \implies x^2-4x-1=0 \implies x = \sqrt{5} \pm 2[/math]

 

[Dr.Peterson]

Notice the RHS can be written as
[math]x = 4 + \dfrac{1}{x} \implies x^2-4x-1=0 \implies x = \sqrt{5} \pm 2[/math]

This is true for the infinite continued fraction, which is probably where they got their problem.

 

[BigBeachBanana]

This is true for the infinite continued fraction, which is probably where they got their problem.

I suspected it was infinite as the LHS and the last denominator have similar forms i.e. [imath]\sqrt{5} + 2[/imath] and [imath] \sqrt{5} -a[/imath].

 

[Steven G]

The 2nd to the last line also has an error. The number of sqrt(5) is not 15, ie 15sqrt(5) is not correct.