![\large\underline{\sf{Solution-}}](https://tex.z-dn.net/?f=%5Clarge%5Cunderline%7B%5Csf%7BSolution-%7D%7D)
We have to find out the value of the fraction.
<u>Let us assume that:</u>
![\sf \longmapsto x =2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ... \infty} } }](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D2%20%2B%20%20%20%5Cdfrac%7B1%7D%7B2%20%2B%20%20%5Cdfrac%7B1%7D%7B2%20%2B%20%20%5Cdfrac%7B1%7D%7B2%20%2B%20...%20%5Cinfty%7D%20%7D%20%7D%20)
<u>We can also write it as:</u>
![\sf \longmapsto x =2 + \dfrac{1}{x}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D2%20%2B%20%5Cdfrac%7B1%7D%7Bx%7D%20)
![\sf \longmapsto x =\dfrac{2x + 1}{x}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%5Cdfrac%7B2x%20%2B%201%7D%7Bx%7D%20)
![\sf \longmapsto {x}^{2} =2x + 1](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%20%7Bx%7D%5E%7B2%7D%20%20%3D2x%20%2B%201)
![\sf \longmapsto {x}^{2} - 2x - 1 = 0](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20%7Bx%7D%5E%7B2%7D%20%20-%202x%20-%201%20%3D%200)
<u>Comparing </u>the given <u>equation</u> with <u>ax² + bx + c = 0,</u> we get:
![\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b = - 2 \\ \sf c = - 1 \end{cases}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%5Cbegin%7Bcases%7D%20%5Csf%20a%20%3D1%20%5C%5C%20%5Csf%20b%20%3D%20%20-%202%20%5C%5C%20%5Csf%20c%20%3D%20%20-%201%20%5Cend%7Bcases%7D)
<u>By quadratic formula:</u>
![\sf \longmapsto x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Cdfrac%7B%20-%20b%20%5Cpm%20%5Csqrt%7B%20%7Bb%7D%5E%7B2%7D%20-%204ac%20%7D%20%7D%7B2a%7D%20)
![\sf \longmapsto x = \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Cdfrac%7B2%20%5Cpm%20%5Csqrt%7B%20%7B%28%20-%202%29%7D%5E%7B2%7D%20-%204%281%29%28%20-%201%29%7D%20%7D%7B2%20%5Ctimes%201%7D%20)
![\sf \longmapsto x = \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Cdfrac%7B2%20%5Cpm%20%5Csqrt%7B4%20%2B%204%7D%20%7D%7B2%20%5Ctimes%201%7D%20)
![\sf \longmapsto x = \dfrac{2 \pm \sqrt{8} }{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Cdfrac%7B2%20%5Cpm%20%5Csqrt%7B8%7D%20%7D%7B2%7D%20)
![\sf \longmapsto x = \dfrac{2 \pm2 \sqrt{2} }{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%20%5Cdfrac%7B2%20%5Cpm2%20%5Csqrt%7B2%7D%20%7D%7B2%7D%20)
![\sf \longmapsto x = 1 \pm\sqrt{2}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%201%20%5Cpm%5Csqrt%7B2%7D)
![\sf \longmapsto x = \begin{cases} \sf 1 + \sqrt{2} \\ \sf 1 - \sqrt{2} \end{cases}](https://tex.z-dn.net/?f=%20%5Csf%20%5Clongmapsto%20x%20%3D%20%5Cbegin%7Bcases%7D%20%5Csf%201%20%20%2B%20%5Csqrt%7B2%7D%20%5C%5C%20%5Csf%201%20-%20%20%5Csqrt%7B2%7D%20%20%5Cend%7Bcases%7D)
<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>
![\sf :\implies x = 1 + \sqrt{2}](https://tex.z-dn.net/?f=%20%5Csf%20%3A%5Cimplies%20x%20%3D%201%20%2B%20%5Csqrt%7B2%7D)
So, the value of the fraction is 1 + √2.
Step-by-step explanation:
5/6+1/4
3/12+10/12=13/12=
well You need to find number which can divide into six and four and the smallest to make finding It easy. 12 is the smallest
If 317+x=609, then x is 292.
609-317=292 therefore, x is 292.
Answer:
g(x) = 1/2x^2
Step-by-step explanation:
Answer:
4:10
24:60
48:120
As long as you multiply or divide both sides of the ratio by the same number, you will have an equivalent ratio.