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Question Number 184438 by Shrinava last updated on 06/Jan/23
If a − (4/( (√a))) = 17 Find a − 4 (√a) = ?
$$\mathrm{If}\:\:\:\mathrm{a}\:−\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{a}}}\:=\:\mathrm{17} \\ $$$$\mathrm{Find}\:\:\:\mathrm{a}\:−\:\mathrm{4}\:\sqrt{\mathrm{a}}\:=\:? \\ $$
Answered by Frix last updated on 06/Jan/23
a−(4/( (√a)))=17 Let t=(√a) ⇒ t>0 t^3 −17t−4=0 (t+4)(t−2−(√5))(t−2+(√5))=0 t>0 ⇒ t=2+(√5) ⇒ a−4(√a)=1
$${a}−\frac{\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{17} \\ $$$$\mathrm{Let}\:{t}=\sqrt{{a}}\:\Rightarrow\:{t}>\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\mathrm{17}{t}−\mathrm{4}=\mathrm{0} \\ $$$$\left({t}+\mathrm{4}\right)\left({t}−\mathrm{2}−\sqrt{\mathrm{5}}\right)\left({t}−\mathrm{2}+\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$${t}>\mathrm{0}\:\Rightarrow\:{t}=\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\Rightarrow \\ $$$${a}−\mathrm{4}\sqrt{{a}}=\mathrm{1} \\ $$
Answered by Frix last updated on 06/Jan/23
a−(4/( (√a)))=17 ⇒ a=(4/( (√a)))+17 (1) a−4(√a)=x ⇒ a=4(√a)+x (2) (2)−(1) ((4a+(x−17)(√a)−4)/( (√a)))=0 4a+(x−17)(√a)−4=0 [a=4(√a)+x] (t+4)(x−1)=0 ⇒ x=1
$${a}−\frac{\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{17}\:\Rightarrow\:{a}=\frac{\mathrm{4}}{\:\sqrt{{a}}}+\mathrm{17}\:\left(\mathrm{1}\right) \\ $$$${a}−\mathrm{4}\sqrt{{a}}={x}\:\Rightarrow\:{a}=\mathrm{4}\sqrt{{a}}+{x}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{4}{a}+\left({x}−\mathrm{17}\right)\sqrt{{a}}−\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{0} \\ $$$$\mathrm{4}{a}+\left({x}−\mathrm{17}\right)\sqrt{{a}}−\mathrm{4}=\mathrm{0} \\ $$$$\left[{a}=\mathrm{4}\sqrt{{a}}+{x}\right] \\ $$$$\left({t}+\mathrm{4}\right)\left({x}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{1} \\ $$
Answered by a.lgnaoui last updated on 06/Jan/23
((a(√a) −4)/( (√a)))=17 a−((4(√a))/a)=17⇒ 4(√a)=a^2 −17a a−4(√a) =18a−a^2 calcul de a 16a=(a^2 −17a)^2 (a^2 −17a−4(√a) )(a^2 −17a+4(√a) )=0 (((√a) )^3 −17((√a) )−4)((√a) )^3 −17(√a) +4)=0 (√a) =z { ((z^3 −17z−4=0)),((z^3 −17z+4=0)) :} .......z? avec a>0 ⇒ a−4(√a) =z^2 −4z
$$\frac{{a}\sqrt{{a}}\:−\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{17} \\ $$$${a}−\frac{\mathrm{4}\sqrt{{a}}}{{a}}=\mathrm{17}\Rightarrow\:\:\:\mathrm{4}\sqrt{{a}}={a}^{\mathrm{2}} −\mathrm{17}{a} \\ $$$${a}−\mathrm{4}\sqrt{{a}}\:=\mathrm{18}{a}−{a}^{\mathrm{2}} \\ $$$${calcul}\:{de}\:{a} \\ $$$$\mathrm{16}{a}=\left({a}^{\mathrm{2}} −\mathrm{17}{a}\right)^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −\mathrm{17}{a}−\mathrm{4}\sqrt{{a}}\:\right)\left({a}^{\mathrm{2}} −\mathrm{17}{a}+\mathrm{4}\sqrt{{a}}\:\right)=\mathrm{0} \\ $$$$\left.\left(\left(\sqrt{{a}}\:\right)^{\mathrm{3}} \:−\mathrm{17}\left(\sqrt{{a}}\:\right)−\mathrm{4}\right)\left(\sqrt{{a}}\:\right)^{\mathrm{3}} \:−\mathrm{17}\sqrt{{a}}\:+\mathrm{4}\right)=\mathrm{0} \\ $$$$\sqrt{{a}}\:={z} \\ $$$$\begin{cases}{{z}^{\mathrm{3}} −\mathrm{17}{z}−\mathrm{4}=\mathrm{0}}\\{{z}^{\mathrm{3}} −\mathrm{17}{z}+\mathrm{4}=\mathrm{0}}\end{cases} \\ $$$$…….{z}?\:\:\:{avec}\:\:{a}>\mathrm{0} \\ $$$$\Rightarrow\:\:\:{a}−\mathrm{4}\sqrt{{a}}\:={z}^{\mathrm{2}} −\mathrm{4}{z} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 06/Jan/23
z^3 −17z−4=0 (z=4,236067) sit a−4(√a) =1 z^3 −17z+4 soit z=4 a−4(√a) =0
$${z}^{\mathrm{3}} −\mathrm{17}{z}−\mathrm{4}=\mathrm{0}\:\left({z}=\mathrm{4},\mathrm{236067}\right) \\ $$$$\:\:\:\:\:\:\:{sit}\:\:{a}−\mathrm{4}\sqrt{{a}}\:=\mathrm{1} \\ $$$${z}^{\mathrm{3}} −\mathrm{17}{z}+\mathrm{4}\:\:\:\:\:\:{soit}\:\:{z}=\mathrm{4} \\ $$$$\:\:\:\:\:\:{a}−\mathrm{4}\sqrt{{a}}\:=\mathrm{0} \\ $$$$ \\ $$
Commented by Frix last updated on 06/Jan/23
z=(√a) ⇔ a=z^2 z=4 ⇒ a=16 a−(4/( (√a)))=16−1=15≠17
$${z}=\sqrt{{a}}\:\Leftrightarrow\:{a}={z}^{\mathrm{2}} \\ $$$${z}=\mathrm{4}\:\Rightarrow\:{a}=\mathrm{16} \\ $$$${a}−\frac{\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{16}−\mathrm{1}=\mathrm{15}\neq\mathrm{17} \\ $$

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