Question-213201

Question Number 213201 by ajfour last updated on 01/Nov/24

Commented by ajfour last updated on 01/Nov/24

$${Find}\:{R} \\ $$

Answered by A5T last updated on 01/Nov/24

(√((R+1)^2 −(R−1)^2 ))=(√(2^2 −1^2 ))+(√((R+1)^2 −R^2 )) ⇒R=5+2(√6)

$$\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −\left({R}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} } \\ $$$$\Rightarrow{R}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$

Commented by A5T last updated on 01/Nov/24

Commented by ajfour last updated on 01/Nov/24

$${Wow},\:{excellent}. \\ $$

Commented by ajfour last updated on 01/Nov/24

(2(√R)−(√3))^2 =1+2R 2R−4(√3)(√R)+2=0 R−2(√3)(√R)+1=0 (√R)=(√3)+(√2) R=5+2(√6)

$$\left(\mathrm{2}\sqrt{{R}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{R} \\ $$$$\mathrm{2}{R}−\mathrm{4}\sqrt{\mathrm{3}}\sqrt{{R}}+\mathrm{2}=\mathrm{0} \\ $$$${R}−\mathrm{2}\sqrt{\mathrm{3}}\sqrt{{R}}+\mathrm{1}=\mathrm{0} \\ $$$$\sqrt{{R}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$$${R}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$

Answered by cherokeesay last updated on 01/Nov/24

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Question-213201

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