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Question-189509




Question Number 189509 by normans last updated on 18/Mar/23
Answered by mr W last updated on 18/Mar/23
Commented by mr W last updated on 18/Mar/23
(xk)^2 =(x−3)k×(2+x+3)k  ⇒x=((15)/2)  (2×1^2 +2^2 +x^2 +(2+x)^2 )k^2 =610  (6+((15^2 )/4)+(((19)/2))^2 )k^2 =610  ⇒k=2  R=(1+x)k     =(1+((15)/2))×2=17
$$\left({xk}\right)^{\mathrm{2}} =\left({x}−\mathrm{3}\right){k}×\left(\mathrm{2}+{x}+\mathrm{3}\right){k} \\ $$$$\Rightarrow{x}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}×\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} +\left(\mathrm{2}+{x}\right)^{\mathrm{2}} \right){k}^{\mathrm{2}} =\mathrm{610} \\ $$$$\left(\mathrm{6}+\frac{\mathrm{15}^{\mathrm{2}} }{\mathrm{4}}+\left(\frac{\mathrm{19}}{\mathrm{2}}\right)^{\mathrm{2}} \right){k}^{\mathrm{2}} =\mathrm{610} \\ $$$$\Rightarrow{k}=\mathrm{2} \\ $$$${R}=\left(\mathrm{1}+{x}\right){k} \\ $$$$\:\:\:=\left(\mathrm{1}+\frac{\mathrm{15}}{\mathrm{2}}\right)×\mathrm{2}=\mathrm{17} \\ $$

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