Value of an indefinitely long expression : Quadratic Equation Approach I came across this problem and have been able to get the correct answer for this although I tried many times on my own.

$$1st$$ Approach : $$8+\frac{8}{8+\frac{1}{8+y}} = y$$ and when I solve this I get to the quadratic equation i.e. $$584 + 7y = 8y^2$$ which does not lead to the correct solution as the roots for this quadratic equation are $$y = \frac{7}{16} \pm \frac{\sqrt(18737)}{16}$$ .

$$2nd$$ Approach : $$8+\frac{8}{8+\frac{1}{y}} = y$$ and when I solve this I get to the quadratic equation i.e. $$8+ 71y = 8y^2$$ which also does not lead to the correct solution as the roots for this quadratic equation are $$y = \frac{71}{16} \pm \frac{\sqrt(5297)}{16}$$ .

PS : This question is a MCQ from where I picked up this question and the options avaliable are

1. $$6\frac{2}{3}$$
2. $$4+\sqrt17$$
3. $$8+\sqrt8$$
4. 9

I am expecting an analytical and step-by-step solution as I was trying to do.

I am assuming that there is a typo in the question as the pattern is not obvious.

Let $$8 + \frac{\color{red}{1}}{8+\frac{1}{8+\ldots}}=y$$

$$8 + \frac1y = y$$ $$y^2-8y-1=0$$

$$y = \frac{8 \pm \sqrt{64+4}}{2}$$

Since $$y > 0$$, $$y = \frac{8 + \sqrt{68}}{2}=4 + \sqrt{17}$$

I did consider other possibilities but I did not hit one of the options: $$8 \left( 1+ \frac1{8 + \frac1{8 + \dots}}\right) = y$$

Now, first, let's study what do we get from $$\frac1{8 + \frac1{8 + \dots}}=z$$

$$\frac1{8+z}=z$$

$$z^2+8z-1=0$$

$$z=\frac{-8 +\sqrt{64+4}}{2}=-4 + \sqrt{17}$$

$$8 (-3+\sqrt{17})=y$$

• yeah..you seems to be reasonably correct...the question might be wrong...thanks for pointing that out ! Apr 11 '21 at 16:24
• Also I have one more question : we can say that the expression given in the original question will be more than 8 but less than 9, right? as the part after the $8+$ will be less than 1 and greater than 0. So with this we can eliminate the options and still get to the right option as only $4+\sqrt17$ is the only option which satisfy the condition. Apr 11 '21 at 16:27
• if $r>0$, we indeed have $8 + \frac{8}{8+r}= 8 + \frac{1}{1+\frac{r}{8}}<9$. The answer should be more than $8$ and less than $9$. Hence, yup, you are right, it seems to be the only option. Apr 11 '21 at 16:33
• thank for confirmation and all the help ! Apr 11 '21 at 16:41