lunpi12
lunpi12
24-03-2017
Mathematics
contestada
Solving radical equation
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jdoe0001
jdoe0001
24-03-2017
[tex]\bf \sqrt{k-9}-\sqrt{k}=-1\implies \sqrt{k-9}=\sqrt{k}-1\impliedby \textit{squaring both sides} \\\\\\ (\sqrt{k-9})^2=(\sqrt{k}-1)^2\implies k-9=(\sqrt{k}-1)(\sqrt{k}-1)\impliedby \textit{FOIL} \\\\\\ k-9=k-2\sqrt{k}+1\implies -10=-2\sqrt{k}\implies \cfrac{-10}{-2}=\sqrt{k} \\\\\\ \sqrt{\cfrac{-10}{-2}}=k[/tex]
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