The value of 3+frac{1}{4+frac{1}{3+frac{1}{4+ | vedclass

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(A) Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$So,$x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{x}{4x+1}$$\Rightarrow x-3=\frac{x}{

Vedclass · Accepted Answer

$1.5+\sqrt{3}$

Vedclass · Answer

(A) Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$So,$x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{x}{4x+1}$$\Rightarrow x-3=\frac{x}{4x+1}$$\Rightarrow (x-3)(4x+1)=x$$\Rightarrow 4x^2+x-12x-3=x$$\Rightarrow 4x^2-12x-3=0$Using the quadratic formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$:$x=\frac{12 \pm \sqrt{(-12)^2-4(4)(-3)}}{2(4)}$$x=\frac{12 \pm \sqrt{144+48}}{8}=\frac{12 \pm \sqrt{192}}{8}$$x=\frac{12 \pm 8\sqrt{3}}{8} = \frac{3}{2} \pm \sqrt{3} = 1.5 \pm \sqrt{3}$Since $x > 0$,we take $x=1.5+\sqrt{3}$.

The value of $3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$ is equal to

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