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Elza [17]
2 years ago
6

The value of

Mathematics
1 answer:
marusya05 [52]2 years ago
4 0

\large\underline{\sf{Solution-}}

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} } }

<u>We can also write it as:</u>

\sf \longmapsto x =2 + \dfrac{1}{x}

\sf \longmapsto x =\dfrac{2x + 1}{x}

\sf \longmapsto  {x}^{2}  =2x + 1

\sf \longmapsto {x}^{2}  - 2x - 1 = 0

<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}

<u>By quadratic formula:</u>

\sf \longmapsto x =  \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}

\sf \longmapsto x =  \dfrac{2 \pm \sqrt{8} }{2}

\sf \longmapsto x =  \dfrac{2 \pm2 \sqrt{2} }{2}

\sf \longmapsto x = 1 \pm\sqrt{2}

\sf \longmapsto x = \begin{cases} \sf 1  + \sqrt{2} \\ \sf 1 -  \sqrt{2}  \end{cases}

<u>But </u><u>"</u><u>x"</u><u> cannot be negative. Therefore:</u>

\sf :\implies x = 1 + \sqrt{2}

So, the value of the fraction is 1 + √2.

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