b- – Tinku Tara

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

\underset{⏟}{b}

Answered by cortano12 last updated on 15/Jul/23

\frac{x - a}{\sqrt{x} + \sqrt{a}} = \frac{x - a}{3 (\sqrt{x} + \sqrt{a})} + 2 \sqrt{a}

\frac{3 (x - a)}{3 (\sqrt{x} + \sqrt{a})} - \frac{(x - a)}{3 (\sqrt{x} + \sqrt{a})} = 2 \sqrt{a}

\frac{2 (x - a)}{3 (\sqrt{x} + \sqrt{a})} = 2 \sqrt{a}

\Rightarrow 2 x - 2 a = 6 \sqrt{ax} + 6 a

\Rightarrow 2 x - 8 a = 6 \sqrt{ax}

\Rightarrow x - 4 a = 3 \sqrt{ax}

\Rightarrow x^{2} - 8 ax + {16 a}^{2} = 9 ax

\Rightarrow x^{2} - 17 ax + {16 a}^{2} = 0

\Rightarrow (x - 16 a) (x - a) = 0, x = a (reject)

\Rightarrow \begin{matrix} x = 16 a \end{matrix}

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

x = a {doesn}^{'} t satisfy in general . It

satisfies only when x = a = 0 .

Commented by cortano12 last updated on 15/Jul/23

yes sir . right

Commented by Frix last updated on 15/Jul/23

x = a = 0 is not allowed because \frac{\dots}{\sqrt{x} + \sqrt{a}} \Rightarrow

\sqrt{x} + \sqrt{a} \neq 0

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

S o r r y f o r m i s t a k e S i r!

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

\underset{―}{⋮ \underset{⏟}{}} x - a

Answered by Frix last updated on 15/Jul/23

x > 0 \land a > 0

\sqrt{x} + \sqrt{a} = t > 0 \Leftrightarrow t - \sqrt{a} > 0 \Leftrightarrow x = {(t - \sqrt{a})}^{2}

Inserting :

t - 2 \sqrt{a} = \frac{t + 4 \sqrt{a}}{3}

t = 5 \sqrt{a}

x = {(5 \sqrt{a} - \sqrt{a})}^{2} = 16 a

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