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Question Number 151307 by mathdanisur last updated on 19/Aug/21

if  x(√x) - 26(√x) = 5  find  x - 5(√x) = ?

$$\mathrm{if}\:\:\mathrm{x}\sqrt{\mathrm{x}}\:-\:\mathrm{26}\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:? \\ $$

Answered by MJS_new last updated on 20/Aug/21

x(√x)−26(√x)−5=0  ((√x)+5)(x−5(√x)−1)=0  ⇒ x−5(√x)=1

$${x}\sqrt{{x}}−\mathrm{26}\sqrt{{x}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\sqrt{{x}}+\mathrm{5}\right)\left({x}−\mathrm{5}\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}−\mathrm{5}\sqrt{{x}}=\mathrm{1} \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Thanmyou Ser

$$\mathrm{Thanmyou}\:\mathrm{Ser} \\ $$

Answered by maged last updated on 20/Aug/21

  if   x(√x)−26(√x)= 5  find  x - 5(√x) = ?  Let :(√x)=u  u^3 −26u=5  u^3 −26u−5=0  u^3 −25u−u−5=0  u(u^2 −25)−(u+5)=0  u(u−5)(u+5)−(u+5)=0  (u+5)[(u(u−5)−1]=0  (u+5)[u^2 −5u−1]=0  ((√x)+5)(x−5(√x)−1)=0  either  (√x) =−5⇒x=25   ⇒ x - 5(√x) = 25−5(−5)=25+25=50  or    x−5(√x)−1=0⇒x−5(√x)=1

$$ \\ $$$$\mathrm{if}\:\:\:\mathrm{x}\sqrt{\mathrm{x}}−\mathrm{26}\sqrt{\mathrm{x}}=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:? \\ $$$$\mathrm{Let}\::\sqrt{\mathrm{x}}=\mathrm{u} \\ $$$$\mathrm{u}^{\mathrm{3}} −\mathrm{26u}=\mathrm{5} \\ $$$$\mathrm{u}^{\mathrm{3}} −\mathrm{26u}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{u}^{\mathrm{3}} −\mathrm{25u}−\mathrm{u}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{u}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{25}\right)−\left(\mathrm{u}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\mathrm{u}\left(\mathrm{u}−\mathrm{5}\right)\left(\mathrm{u}+\mathrm{5}\right)−\left(\mathrm{u}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left(\mathrm{u}+\mathrm{5}\right)\left[\left(\mathrm{u}\left(\mathrm{u}−\mathrm{5}\right)−\mathrm{1}\right]=\mathrm{0}\right. \\ $$$$\left(\mathrm{u}+\mathrm{5}\right)\left[\mathrm{u}^{\mathrm{2}} −\mathrm{5u}−\mathrm{1}\right]=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{x}}+\mathrm{5}\right)\left(\mathrm{x}−\mathrm{5}\sqrt{\mathrm{x}}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{either} \\ $$$$\sqrt{\mathrm{x}}\:=−\mathrm{5}\Rightarrow\mathrm{x}=\mathrm{25} \\ $$$$\:\Rightarrow\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:\mathrm{25}−\mathrm{5}\left(−\mathrm{5}\right)=\mathrm{25}+\mathrm{25}=\mathrm{50} \\ $$$$\mathrm{or} \\ $$$$\:\:\mathrm{x}−\mathrm{5}\sqrt{\mathrm{x}}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}−\mathrm{5}\sqrt{\mathrm{x}}=\mathrm{1} \\ $$$$ \\ $$

Commented by MJS_new last updated on 20/Aug/21

why you solve  26(√x)−x(√x)=5  when the given equation is  x(√x)−26(√x)=5  ???

$$\mathrm{why}\:\mathrm{you}\:\mathrm{solve} \\ $$$$\mathrm{26}\sqrt{{x}}−{x}\sqrt{{x}}=\mathrm{5} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{is} \\ $$$${x}\sqrt{{x}}−\mathrm{26}\sqrt{{x}}=\mathrm{5} \\ $$$$??? \\ $$

Commented by mathdanisur last updated on 20/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

Commented by MJS_new last updated on 20/Aug/21

(√x)=−5 has no solution

$$\sqrt{{x}}=−\mathrm{5}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution} \\ $$

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