wrong solution (Question:837672.0)

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Uttam singh

· started a discussion

· 1 Months ago

wrong solution (Question:837672.0)

wrong solution

Question:

If x = $\frac{1}{3 + \frac{1}{4 + \frac{1}{3 + \frac{1}{4 + . . . . . . . . \infty}}}}$ ,What is the value of x ?

Options:

$\frac{- 6 \pm 2 \sqrt{6}}{3}$

$\frac{- 6 \pm \sqrt{15}}{6}$

$\frac{- 3 \pm \sqrt{15}}{4}$

$\frac{- 6 \pm \sqrt{24}}{6}$

Solution:

x = $\frac{1}{3 + \frac{1}{4 + x}}$

x = $\frac{4 + x}{12 + 3 x + 1}$

12x + 3x² + x = 4 + x

3x² + 12x – 4 = 0

x = $\frac{- 12 \pm \sqrt{144 - 4 \times 4 \times 3}}{6}$

= $\frac{- 12 \pm \sqrt{144 - 48}}{6}$

= $\frac{- 12 \pm 4 \sqrt{6}}{6}$

= $\frac{- 6 \pm 2 \sqrt{6}}{3}$

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wrong solution (Question:837672.0)

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