Question

please answer
what is p equal to?​

Answer 1

Answer:

hey here is your solution

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Step-by-step explanation:

$to \: find = \\ value \: of \: p \\ \\ so \: here \\ \frac{a + p + \sqrt{a {}^{2} -p {}^{2} } }{a + p - \sqrt{a {}^{2} - p {}^{2} } } = \frac{b}{p} \\ \\ applying \: componendo \: dividendo \\ we \: get \\ \\ = \frac{a + p + \sqrt{a {}^{2} - p {}^{2} } \: + \: a + p - \sqrt{a {}^{2} - p {}^{2} } }{a +p + \sqrt{a {}^{2} - p {}^{2} } \: - \: (a + p - \sqrt{a {}^{2} - p {}^{2} } \: ) } = \frac{b + p}{b - p} \\ \\ = \frac{a + p + a + p}{a + p + \sqrt{a {}^{2} - p {}^{2} } - a - p + \sqrt{a {}^{2} - p {}^{2} } } = \frac{b + p}{b - p} \\ \\ = \frac{2a + 2p}{ \sqrt{a {}^{2} - p {}^{2} } \: + \: \sqrt{a {}^{2} - p {}^{2} } } = \frac{b + p}{b - p} \\ \\ = \frac{2(a + p)}{2. \sqrt{a {}^{2} - p {}^{2} } } = \frac{b + p}{b - p} \\ \\ = \frac{a + p}{ \sqrt{a {}^{2} - p {}^{2} } } = \frac{b + p}{b - p} \\ \\ squaring \: both \: sides \\ we \: get \\ \\ = \frac{(a + p) {}^{2} }{( \sqrt{a {}^{2} - p {}^{2} } \: )} = \frac{(b + p) {}^{2} }{(b - p) {}^{2} } \\ \\ = \frac{(a + p)(a + p)}{a {}^{2} - p {}^{2} } \: = \frac{b {}^{2} + p {}^{2} + 2bp}{b {}^{2} + p {}^{2} - 2bp } \\ \\ = \frac{(a + p)(a + p)}{(a + p)(a - p)} = \frac{b {}^{2} + p {}^{2} + 2bp }{b {}^{2} + p {}^{2} - 2bp } \\ \\ = \frac{a + p}{a - p} = \frac{b {}^{2} + p {}^{2} + 2bp}{b {}^{2} + p {}^{2} - 2bp}$

$again \: applying \: compenendo \: dividendo \\ we \: get \\ \\ = \frac{a + p + a - p}{a + p - (a - p)} = \frac{b {}^{2} + p {}^{2} + bp + b {}^{2} + p {}^{2} - 2bp}{b {}^{2} + p {}^{2} + 2bp - (b {}^{2} + p {}^{2} - 2bp) } \\ \\ = \frac{a + a}{a + p - a + p} = \frac{b {}^{2} + p {}^{2} + b {}^{2} + p {}^{2} }{b {}^{2} + p {}^{2} + 2bp - b {}^{2} - p {}^{2} + 2bp } \\ \\ = \frac{2a}{2p} = \frac{2b {}^{2} + 2p {}^{2} }{2bp + 2bp} \\ \\ = \frac{a}{p} = \frac{2(b {}^{2} + p {}^{2} )}{4bp} \\ \\ = a = \frac{b {}^{2} + p {}^{2} }{2b} \\ \\ = 2ab = b {}^{2} + p {}^{2} \\ = p {}^{2} = 2ab - b {}^{2} \\ = b(2a - b) \\ \\ taking \: square \: root \: on \: both \: sides$

$we \: get \\ p = ± \: \sqrt{b(2a - b)}$