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Question-174487




Question Number 174487 by mnjuly1970 last updated on 02/Aug/22
Answered by dragan91 last updated on 02/Aug/22
  x^2 −25x=x+5(√x)  x(x−25)=(√x)((√x)+5)  x((√x)−5)((√x)+5)=(√x)((√x)+5)  x((√x)−5)=(√x)/:(√x)  (√x)((√x)−5)=1  x−5(√x)=1/.5  5x−25(√x)=5 for x≠0
$$ \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{25x}=\mathrm{x}+\mathrm{5}\sqrt{\mathrm{x}} \\ $$$$\mathrm{x}\left(\mathrm{x}−\mathrm{25}\right)=\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}+\mathrm{5}\right) \\ $$$$\mathrm{x}\left(\sqrt{\mathrm{x}}−\mathrm{5}\right)\left(\sqrt{\mathrm{x}}+\mathrm{5}\right)=\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}+\mathrm{5}\right) \\ $$$$\mathrm{x}\left(\sqrt{\mathrm{x}}−\mathrm{5}\right)=\sqrt{\mathrm{x}}/:\sqrt{\mathrm{x}} \\ $$$$\sqrt{\mathrm{x}}\left(\sqrt{\mathrm{x}}−\mathrm{5}\right)=\mathrm{1} \\ $$$$\mathrm{x}−\mathrm{5}\sqrt{\mathrm{x}}=\mathrm{1}/.\mathrm{5} \\ $$$$\mathrm{5x}−\mathrm{25}\sqrt{\mathrm{x}}=\mathrm{5}\:\mathrm{for}\:\mathrm{x}\neq\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 03/Aug/22
very nice solution thx alot   sir dragan
$${very}\:{nice}\:{solution}\:{thx}\:{alot}\: \\ $$$${sir}\:{dragan} \\ $$

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