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Question Number 149781 by mathdanisur last updated on 07/Aug/21

a − (√((20)/a)) = 7 ⇒ (√(5a)) − a = ?

$$\mathrm{a}\:\:−\:\:\sqrt{\frac{\mathrm{20}}{\mathrm{a}}}\:=\:\mathrm{7}\:\:\Rightarrow\:\:\sqrt{\mathrm{5a}}\:−\:\mathrm{a}\:=\:? \\ $$

Commented by amin96 last updated on 07/Aug/21

(√((20)/a))=a−7 ⇒ (√(5a))=((a^2 −7a)/2) (√(5a))−a=((5a−a^2 )/( (√(5a))+a))=((5a−a^2 )/(((a^2 −7a)/2)+a))=((5a−a^2 )/((a^2 −5a)/2))=−2

$$\sqrt{\frac{\mathrm{20}}{{a}}}={a}−\mathrm{7}\:\:\:\Rightarrow\:\:\sqrt{\mathrm{5}{a}}=\frac{{a}^{\mathrm{2}} −\mathrm{7}{a}}{\mathrm{2}}\:\: \\ $$$$\sqrt{\mathrm{5}{a}}−{a}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{5}{a}}+{a}}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}} −\mathrm{7}{a}}{\mathrm{2}}+{a}}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}} −\mathrm{5}{a}}{\mathrm{2}}}=−\mathrm{2} \\ $$$$ \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Answered by MJS_new last updated on 07/Aug/21

you can easily do it: transform to (√((20)/a))=a−7 square it, solve the 3^(rd) degree polynome (hint: one solution is a=5), test all solutions if they fit the given equation and you′re done

$$\mathrm{you}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{do}\:\mathrm{it}: \\ $$$$\mathrm{transform}\:\mathrm{to}\:\sqrt{\frac{\mathrm{20}}{{a}}}={a}−\mathrm{7} \\ $$$$\mathrm{square}\:\mathrm{it},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}\:\mathrm{polynome} \\ $$$$\left(\mathrm{hint}:\:\mathrm{one}\:\mathrm{solution}\:\mathrm{is}\:{a}=\mathrm{5}\right),\:\mathrm{test}\:\mathrm{all}\:\mathrm{solutions} \\ $$$$\mathrm{if}\:\mathrm{they}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{you}'\mathrm{re}\:\mathrm{done} \\ $$

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