Question

p=2+root3 find p+1/p and p^2+1/p^2

Answer 1

Answer:

here is yur answer bro or sis

Answer 2

Question :-

$\rm{p = 2 + \sqrt{3} }$ find $\rm{p + \frac{1}{p} }$ and $\rm{p {}^{2} + \frac{1}{p {}^{2} } }$ .

Answer :-

$⏩ \: \rm{p + \frac{1}{p} = 4 }$

$⏩ \: \rm{p {}^{2} + \frac{1}{p {}^{2} } = 14 }$

Step by step explanation :-

Given that,

$\rm\implies{p = 2 + \sqrt{3} }$

Now,

$\rm:\longmapsto \red{p = 2 + \sqrt{3} } \: \: \: \\ \\ \rm:\longmapsto{ \frac{1}{p} = \frac{1}{2 + \sqrt{3} } }$

On Rationalizing the denominator we get,

$\rm:\longmapsto{ \frac{1}{p} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } } \\ \\ \rm:\longmapsto{ \frac{1}{p} = \frac{2 - \sqrt{3} }{(2 + \sqrt{3)} (2 - \sqrt{3}) } }$

In denominator using (a+b) (a-b) = a² - b², On using this identity we get,

$\rm:\longmapsto{ \frac{1}{p} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3} ) {}^{2} } } \\ \\ \rm:\longmapsto{ \frac{1}{p} = \frac{2 - \sqrt{3} }{1} } \: \: \: \: \: \: \: \: \:$

Hence,

$\rm:\longmapsto \red{ \frac{1}{p} = 2 - \sqrt{3} }$

We have to find,

$\rm:\longmapsto{p + \frac{1}{p} }$

On substituting all values we get,

$\rm:\longmapsto{p + \frac{1}{p} = 2 + \cancel{ \sqrt{3} }+ 2 - \cancel{ \sqrt{3} }} \\ \\ \rm:\longmapsto{p + \frac{1}{p} = 2 + 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto \red{p + \frac{1}{p} = 4 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

In 2nd question we have to find,

$\rm:\longmapsto{p {}^{2} + \frac{1}{p {}^{2} } }$

On substituting all values we get,

$\rm:\longmapsto{p {}^{2} + \frac{1}{p {}^{2} } = (2 + \sqrt{3} ) {}^{2} + (2 - \sqrt{3} ) {}^{2} }$

We using,

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² - 2ab + b²

On using this identity we get,

$\small\rm:\longmapsto{p {}^{2} + \frac{1}{p {}^{2} } = \{ (2) {}^{2} + 2(2)( \sqrt{3} ) + ( \sqrt{3}) {}^{2} \} + \{(2) {}^{2} - 2(2)( \sqrt{3}) + ( \sqrt{3} ) {}^{2} } \\ \\ \small\rm:\longmapsto{p {}^{2} + \frac{1}{p {}^{2} } = 4 \: \: \cancel{+ 4 \sqrt{3} } + 3 + 4 \: \: \cancel{- 4 \sqrt{3} }+ 3 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto{p {}^{2} + \frac{1}{p {}^{2} } = 4 + 3 + 4 + 3 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \small\rm:\longmapsto \red{p {}^{2} + \frac{1}{p {}^{2} } = 14} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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