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Question-122203




Question Number 122203 by help last updated on 14/Nov/20
Commented by Dwaipayan Shikari last updated on 14/Nov/20
K^2 x+8y=30  Kx+8y=10  x(K^2 −K)=20⇒K^2 =((20)/x)+K  K=((1+(√(1+((80)/x))))/2)  x=1  K=5( Not satisfies)  x=10   K=2    (Satisfies)  So K^2 =4
$${K}^{\mathrm{2}} {x}+\mathrm{8}{y}=\mathrm{30} \\ $$$${Kx}+\mathrm{8}{y}=\mathrm{10} \\ $$$${x}\left({K}^{\mathrm{2}} −{K}\right)=\mathrm{20}\Rightarrow{K}^{\mathrm{2}} =\frac{\mathrm{20}}{{x}}+{K} \\ $$$${K}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{80}}{{x}}}}{\mathrm{2}} \\ $$$${x}=\mathrm{1}\:\:{K}=\mathrm{5}\left(\:{Not}\:{satisfies}\right) \\ $$$${x}=\mathrm{10}\:\:\:{K}=\mathrm{2}\:\:\:\:\left({Satisfies}\right) \\ $$$${So}\:{K}^{\mathrm{2}} =\mathrm{4} \\ $$

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