Question

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Answer 1

Answer:

$Let \: p(x) = {x}^{2} - (p + \frac{1}{p} )x + 1 \\ p(x) = 0 \\ {x}^{2} - (\frac{ {p}^{2} + 1 }{p})x + 1 = 0 \\ {x}^{2} - ( \frac{ {p}^{2}x + x }{p} ) + 1 = 0 \\ Multiply \: both \: sides \: by \: p \\ p({x}^{2} - ( \frac{ {p}^{2}x + x }{p} ) + 1) = p(0 ) \\ p {x}^{2} - \cancel p ( \frac{ {p}^{2}x + x }{ \cancel p} ) + p = 0 \\ p {x}^{2} - ( {p}^{2} x + x) + p = 0 \\ p {x}^{2} - {p}^{2}x - x + p = 0 \\ px(x - p) - 1(x - p) = 0 \\ (px - 1)(x - p) = 0 \\ px - 1 = 0 \: ,x - p = 0 \\ px = 1,x = p \\ \boxed{x = \frac{1}{p}} , \boxed{x = p}$

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Answer 2

Answer:

$Let \: p(x) = {x}^{2} - (p + \frac{1}{p} )x + 1 \\ p(x) = 0 \\ {x}^{2} - (\frac{ {p}^{2} + 1 }{p})x + 1 = 0 \\ {x}^{2} - ( \frac{ {p}^{2}x + x }{p} ) + 1 = 0 \\ Multiply \: both \: sides \: by \: p \\ p({x}^{2} - ( \frac{ {p}^{2}x + x }{p} ) + 1) = p(0 ) \\ p {x}^{2} - \cancel p ( \frac{ {p}^{2}x + x }{ \cancel p} ) + p = 0 \\ p {x}^{2} - ( {p}^{2} x + x) + p = 0 \\ p {x}^{2} - {p}^{2}x - x + p = 0 \\ px(x - p) - 1(x - p) = 0 \\ (px - 1)(x - p) = 0 \\ px - 1 = 0 \: ,x - p = 0 \\ px = 1,x = p \\ \boxed{x = \frac{1}{p}} , \boxed{x = p}$

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