If-P-2-1-P-then-what-is-the-answer-of-P-2-1-P-2-

Question Number 134135 by Agnibhoo last updated on 28/Feb/21

If P = 2 + \frac{1}{P} then what is the answer of

P^{2} - \frac{1}{P^{2}} ?

Answered by Ñï= last updated on 28/Feb/21

p^{2} - \frac{1}{p^{2}}

= (p - \frac{1}{p}) (p + \frac{1}{p})

= (p - \frac{1}{p}) {[{(p - \frac{1}{p})}^{2} + 4]}^{1 / 2}

= 2 \times {(2^{2} + 4)}^{1 / 2}

= 4 \sqrt{2}

Answered by bobhans last updated on 28/Feb/21

p - \frac{1}{p} = 2 \Rightarrow p^{2} + \frac{1}{p^{2}} = 6

\Rightarrow {(p + \frac{1}{p})}^{2} - 2 = 6 \Rightarrow p + \frac{1}{p} = 2 \sqrt{2}

t h e n (p - \frac{1}{p}) (p + \frac{1}{p}) = 2 \times 2 \sqrt{2}

p^{2} - \frac{1}{p^{2}} = 4 \sqrt{2}

Answered by mr W last updated on 28/Feb/21

P - \frac{1}{P} = 2

P^{2} - 2 + \frac{1}{P^{2}} = 4

P^{2} + 2 + \frac{1}{P^{2}} = 8

{(P + \frac{1}{P})}^{2} = 8

P + \frac{1}{P} = \pm 2 \sqrt{2}

P^{2} - \frac{1}{P^{2}} = (P + \frac{1}{P}) (P - \frac{1}{P}) = \pm 4 \sqrt{2}

Answered by Rasheed.Sindhi last updated on 28/Feb/21

P = 2 + \frac{1}{P} \Rightarrow p^{2} - 2 p - 1 = 0

p = 1 \pm \sqrt{2} \Rightarrow p^{2} = 1 + 2 \pm 2 \sqrt{2} = 3 \pm 2 \sqrt{2}

p^{2} - \frac{1}{p^{2}} = (3 \pm 2 \sqrt{2}) - \frac{1}{3 \pm 2 \sqrt{2}} . \frac{3 \mp 2 \sqrt{2}}{3 \mp 2 \sqrt{2}}

= 3 \pm 2 \sqrt{2} - \frac{3 \mp 2 \sqrt{2}}{9 - 8} = 3 \pm 2 \sqrt{2} - 3 \pm 2 \sqrt{2}

= \pm 4 \sqrt{2}