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

If x = (√(42−(√(42−(√(42−...))))))  y = (√(x+(√(x+(√(x+...))))))  z=(√(y.(√(y.(√(y.(√(y...)))))))) . Find x+y+z .

$${If}\:{x}\:=\:\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−...}}} \\ $$$${y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}} \\ $$$${z}=\sqrt{{y}.\sqrt{{y}.\sqrt{{y}.\sqrt{{y}...}}}}\:.\:{Find}\:{x}+{y}+{z}\:. \\ $$

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

x^2 =42−(√(42−(√(42−...))))=42−x ⇒x^2 +x−42 =0     Δ=1−4(−42) =1+4.42 =169 ⇒x_1 =((−1+13)/2)=6  x_2 =((−1−13)/2)=−7 (x_2 <0 to eliminate) ⇒x=6  y^2 =x+y ⇒y^2 −y−6 =0  Δ=1−4(−6) =25 ⇒y_1 =((1+5)/2)=3  and y_2 =((1−5)/2)=−2 <0 to eliminate  ⇒y=3  ⇒z^2  =yz ⇒z =y=3 ⇒x+y+z =6+3+3 =12

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

Commented by bemath last updated on 16/Oct/20

thank you

$${thank}\:{you} \\ $$

Commented by Bird last updated on 16/Oct/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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