Let 6+6+6+...=k6+k=k6+k=k2⇒k2−k−6=0⇒k2−3k+2k−6=0⇒k(k−3)+2(k−3)=0⇒(k−3)(k+2)=0∴k=3 or −2

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$\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}}$ equal is

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$6^{\frac{1}{3}}$

$6^{\frac{2}{3}}$

answer is D.

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Detailed Solution

Let $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = k$

$\sqrt{6 + k} = k$

$6 + k = k^{2} \Rightarrow k^{2} - k - 6 = 0$

$\Rightarrow k^{2} - 3 k + 2 k - 6 = 0$

$\Rightarrow k (k - 3) + 2 (k - 3) = 0$

$\Rightarrow (k - 3) (k + 2) = 0$

$∴ k = 3$ or $- 2$

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