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Question Number 194756 by horsebrand11 last updated on 15/Jul/23

$$\:\:\:\:\:\:\underbrace{\boldsymbol{{b}}} \\ $$

Answered by cortano12 last updated on 15/Jul/23

((x−a)/( (√x)+(√a))) = ((x−a)/(3((√x)+(√a)))) +2(√a) ((3(x−a))/(3((√x)+(√a))))−(((x−a))/(3((√x)+(√a)))) = 2(√a) ((2(x−a))/(3((√x)+(√a)))) = 2(√a) ⇒2x−2a = 6(√(ax))+6a ⇒2x−8a = 6(√(ax)) ⇒x−4a = 3(√(ax)) ⇒x^2 −8ax+16a^2 = 9ax ⇒x^2 −17ax+16a^2 =0 ⇒(x−16a)(x−a)=0, x=a (reject) ⇒ determinant (((x=16a)))

$$\:\:\frac{\mathrm{x}−\mathrm{a}}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}}\:=\:\frac{\mathrm{x}−\mathrm{a}}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:+\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\frac{\mathrm{3}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}−\frac{\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\:\frac{\mathrm{2}\left(\mathrm{x}−\mathrm{a}\right)}{\mathrm{3}\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{a}}\right)}\:=\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2x}−\mathrm{2a}\:=\:\mathrm{6}\sqrt{\mathrm{ax}}+\mathrm{6a} \\ $$$$\:\:\:\:\Rightarrow\mathrm{2x}−\mathrm{8a}\:=\:\mathrm{6}\sqrt{\mathrm{ax}} \\ $$$$\:\:\:\Rightarrow\mathrm{x}−\mathrm{4a}\:=\:\mathrm{3}\sqrt{\mathrm{ax}} \\ $$$$\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{8ax}+\mathrm{16a}^{\mathrm{2}} =\:\mathrm{9ax} \\ $$$$\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{17ax}+\mathrm{16a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\Rightarrow\left(\mathrm{x}−\mathrm{16a}\right)\left(\mathrm{x}−\mathrm{a}\right)=\mathrm{0},\:\mathrm{x}=\mathrm{a}\:\left(\mathrm{reject}\right) \\ $$$$\:\:\:\Rightarrow\:\begin{array}{|c|}{\mathrm{x}=\mathrm{16a}}\\\hline\end{array} \\ $$$$\:\:\: \\ $$

Commented by Rasheed.Sindhi last updated on 15/Jul/23

x=a doesn′t satisfy in general.It satisfies only when x=a=0.

$$\mathrm{x}=\mathrm{a}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{satisfy}\:\mathrm{in}\:\mathrm{general}.\mathrm{It} \\ $$$$\mathrm{satisfies}\:\mathrm{only}\:\mathrm{when}\:\mathrm{x}=\mathrm{a}=\mathrm{0}. \\ $$

Commented by cortano12 last updated on 15/Jul/23

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{right}\: \\ $$

Commented by Frix last updated on 15/Jul/23

x=a=0 is not allowed because ((...)/( (√x)+(√a))) ⇒ (√x)+(√a)≠0

$${x}={a}=\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{allowed}\:\mathrm{because}\:\frac{…}{\:\sqrt{{x}}+\sqrt{{a}}}\:\Rightarrow \\ $$$$\sqrt{{x}}+\sqrt{{a}}\neq\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 15/Jul/23

$${Sorry}\:{for}\:{mistake}\:{Sir}! \\ $$

Answered by Rasheed.Sindhi last updated on 15/Jul/23

$$\:\underline{\vdots\cancel{\underbrace{ }}}\mathrm{x}−\mathrm{a} \\ $$

Answered by Frix last updated on 15/Jul/23

x>0∧a>0 (√x)+(√a)=t>0 ⇔ t−(√a)>0 ⇔ x=(t−(√a))^2 Inserting: t−2(√a)=((t+4(√a))/3) t=5(√a) x=(5(√a)−(√a))^2 =16a

$${x}>\mathrm{0}\wedge{a}>\mathrm{0} \\ $$$$\sqrt{{x}}+\sqrt{{a}}={t}>\mathrm{0}\:\Leftrightarrow\:{t}−\sqrt{{a}}>\mathrm{0}\:\Leftrightarrow\:{x}=\left({t}−\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$\mathrm{Inserting}: \\ $$$${t}−\mathrm{2}\sqrt{{a}}=\frac{{t}+\mathrm{4}\sqrt{{a}}}{\mathrm{3}} \\ $$$${t}=\mathrm{5}\sqrt{{a}} \\ $$$${x}=\left(\mathrm{5}\sqrt{{a}}−\sqrt{{a}}\right)^{\mathrm{2}} =\mathrm{16}{a} \\ $$

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