Question

find X {√(a+x)+√(a-x)}/{√(a+x)-√(a-x)}=c/d​

Answer 1

Answer:

$\large{ \bold{x = 0 \: , \: \frac{2acd}{ {c}^{2} + {d}^{2} }}}$

Step-by-step explanation:

$\frac{ \sqrt{a + x} + \sqrt{a - x} }{\sqrt{a + x} - \sqrt{a - x}} = \frac{c}{d} \\ \\ = > \frac{c}{d} = \frac{(\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a + x} - \sqrt{a - x})} \times \frac{(\sqrt{a + x} + \sqrt{a - x})}{(\sqrt{a + x} + \sqrt{a - x})} \\ \\ = > \frac{c}{d} = \frac{ {(\sqrt{a + x} + \sqrt{a - x})}^{2} }{ {( \sqrt{a + x} )}^{2} - {( \sqrt{a - x}) }^{2} } \\ \\ = > \frac{c}{d} = \frac{2a + 2 \sqrt{ {a}^{2} - {x}^{2} } }{2x} \\ \\ = > \frac{c}{d} = \frac{a + \sqrt{ {a}^{2} - {x}^{2} } }{x} \\ \\ = > xc - da = d \sqrt{ {a}^{2} - {x}^{2} } \\ \\ sobs \\ \\ = > {c}^{2} {x}^{2} + {a}^{2} {d}^{2} - 2acdx = {a}^{2} {d}^{2} - {d}^{2} {x}^{2} \\ \\ = > ( {c}^{2} + {d}^{2} ) {x}^{2} - 2acdx = 0 \\ \\ = > x(( {c}^{2} + {d}^{2} )x - 2acd) = 0 \\ \\ = > x = 0 \\ \\ = > x = \frac{2acd}{ {c}^{2} + {d}^{2} }$

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