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Question Number 118227 by bemath last updated on 16/Oct/20

$$Ifx=\sqrt{42-\sqrt{42-\sqrt{42-...}}}$$$$y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}$$$$z=\sqrt{y.\sqrt{y.\sqrt{y.\sqrt{y...}}}}.Findx+y+z.$$

Answered by mathmax by abdo last updated on 16/Oct/20

$${\mathrm{x}}^2=42-\sqrt{42-\sqrt{42-...}}=42-\mathrm{x}\Rightarrow {\mathrm{x}}^2+\mathrm{x}-42=0$$$$\mathrm{\Delta }=1-4(-42)=1+4.42=169\Rightarrow {\mathrm{x}}_1=\frac{-1+13}{2}=6$$$${\mathrm{x}}_2=\frac{-1-13}{2}=-7({\mathrm{x}}_2<0\mathrm{to}\mathrm{eliminate})\Rightarrow \mathrm{x}=6$$$${\mathrm{y}}^2=\mathrm{x}+\mathrm{y}\Rightarrow {\mathrm{y}}^2-\mathrm{y}-6=0$$$$\mathrm{\Delta }=1-4(-6)=25\Rightarrow {\mathrm{y}}_1=\frac{1+5}{2}=3\mathrm{and}{\mathrm{y}}_2=\frac{1-5}{2}=-2<0\mathrm{to}\mathrm{eliminate}$$$$\Rightarrow \mathrm{y}=3\Rightarrow {\mathrm{z}}^2=\mathrm{yz}\Rightarrow \mathrm{z}=\mathrm{y}=3\Rightarrow \mathrm{x}+\mathrm{y}+\mathrm{z}=6+3+3=12$$

Commented by bemath last updated on 16/Oct/20

$$thankyou$$

Commented by Bird last updated on 16/Oct/20

$$youarewelcome$$

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