Question

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Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

We have to find out the value of the fraction.

Let us assume that:

$\sf \implies x =2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ... \infty} } }$

We can also write it as:

$\sf \implies x =2 + \dfrac{1}{x}$

$\sf \implies x =\dfrac{2x + 1}{x}$

$\sf \implies {x}^{2} =2x + 1$

$\sf \implies {x}^{2} - 2x - 1 = 0$

Comparing the given equation with ax² + bx + c = 0, we get:

$\sf \implies \begin{cases} \sf a =1 \\ \sf b = - 2 \\ \sf c = - 1 \end{cases}$

By quadratic formula:

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

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

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

$\sf \implies x = \dfrac{2 \pm \sqrt{8} }{2}$

$\sf \implies x = \dfrac{2 \pm2 \sqrt{2} }{2}$

$\sf \implies x = 1 \pm\sqrt{2}$

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

But x cannot be negative. Therefore:

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

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