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Question Number 217842 by aylinrabbani last updated on 22/Mar/25

f(x)=(√(x^2 −6x+10)) f(3+2(√6))=?

$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{10}} \\ $$$${f}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right)=? \\ $$

Answered by Rasheed.Sindhi last updated on 22/Mar/25

f(x)=(√(x^2 −6x+10)) =(√(x^2 −6x+9+1)) =(√((x−3)^2 +1)) f(3+2(√6))=(√((3+2(√6) −3)^2 +1)) =(√((2(√6) )^2 +1)) =(√(4(6)+1)) =(√(25)) =5

$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}}\: \\ $$$${f}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right)=\sqrt{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\:−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\left(\mathrm{2}\sqrt{\mathrm{6}}\:\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{4}\left(\mathrm{6}\right)+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{25}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{5} \\ $$

Answered by AntonCWX8 last updated on 22/Mar/25

(3+2(√6))^2 −6(3+2(√6))+10 =9+12(√6)+24−18−12(√6)+10 =25 (√(25))=±5

$$\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} −\mathrm{6}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right)+\mathrm{10} \\ $$$$=\mathrm{9}+\mathrm{12}\sqrt{\mathrm{6}}+\mathrm{24}−\mathrm{18}−\mathrm{12}\sqrt{\mathrm{6}}+\mathrm{10} \\ $$$$=\mathrm{25} \\ $$$$ \\ $$$$\sqrt{\mathrm{25}}=\pm\mathrm{5} \\ $$

Commented by mr W last updated on 22/Mar/25

(√(25))=5≠±5

$$\sqrt{\mathrm{25}}=\mathrm{5}\neq\pm\mathrm{5} \\ $$

Commented by Frix last updated on 22/Mar/25

If (√(25))=±5 then (√(25))+(√(25))=? (a) −10 (b) 0 (c) 10

$$\mathrm{If}\:\sqrt{\mathrm{25}}=\pm\mathrm{5}\:\mathrm{then}\:\sqrt{\mathrm{25}}+\sqrt{\mathrm{25}}=? \\ $$$$\left(\mathrm{a}\right)\:−\mathrm{10} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{0} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{10} \\ $$

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