<img src="https://tex.z-dn.net/?f=%5Csqrt%7B4%2B%5Csqrt%7B4%2B%5Csqrt%7B4%2B...%2B%5Csqrt%7B4%7D" id="TexFormula1" title="\sqrt{

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torisob [31]

3 years ago

4+\sqrt{4+\sqrt{4+...+\sqrt{4}" alt="\sqrt{4+\sqrt{4+\sqrt{4+...+\sqrt{4}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Sonbull [250]3 years ago

8 0

Answer:

y=0.5+sqrt(17)

Step-by-step explanation:

Let y=sqrt{4+sqrt{4+...+sqrt(4)}

y=sqrt(4+y). (Since it's an infinite series)

y^2=4+y, y=0.5-sqrt(17) or 0.5+sqrt(17). We will omit the negative since root values are positive.

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