Question

solve it with solution​

Answer 1

Given to solve:-

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$\dfrac {3a}{\sqrt{x+a}}=\sqrt x+\sqrt{x+a}$

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Let,

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$k=\sqrt {x+a}$

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Then,

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$\dfrac {3a}{k}=\sqrt x+k\\\\\\k^2+k\sqrt x-3a=0$

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Solving the quadratic equation,

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$k=\dfrac {-\sqrt x\pm\sqrt{x+12a}}{2}\\\\\\2\sqrt{x+a}=-\sqrt x\pm\sqrt{x+12a}$

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Squaring both sides,

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$4(x+a)=2x+12a\pm\sqrt {x^2+12ax}\\\\\\2x-8a=\pm\sqrt {x^2+12ax}$

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Again squaring both the sides,

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$4x^2+64a^2-32ax=x^2+12ax\\\\\\3x^2-44ax+64a^2=0\\\\\\3x^2+(\sqrt {292}-22)ax-(\sqrt {292}+22)ax+64a^2=0\\\\\\x\Bigg(3x+(\sqrt {292}-22)a\Bigg)-\dfrac {(\sqrt {292}+22)a}{3}\Bigg(3x+(\sqrt {292}-22)a\Bigg)=0\\\\\\\Bigg(3x+(\sqrt {292}-22)a\Bigg)\left(x-\dfrac {(\sqrt {292}+22)a}{3}\right)=0$

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Thus,

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$3x-(22-\sqrt {292})a=0\quad\quad OR\quad\quad 3x-(22+\sqrt {292})a=0$$\quad$

Or simply,

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$\Large\text { \underline {\underline {3x-(22\pm\sqrt {292})a=0}} }$