Question

If p=2-root5/2+root5 and q=2+root5/2-root5 then find p^2+q^2 and p^2-q^2.​

Answer 1

Required Answer:-

Given:

• p = (2 - √5)/(2 + √5)
• q = (2 + √5)(2 - √5)

To Find:

• p² - q² = ?

Solution:

Given,

$\sf \implies p = \dfrac{2 - \sqrt{5} }{2 + \sqrt{5} }$

$\sf \implies p = \dfrac{(2 - \sqrt{5} )}{(2 + \sqrt{5}) } \times \dfrac{(2 - \sqrt{5} )}{(2 - \sqrt{5})}$

$\sf \implies p = \dfrac{(2 - \sqrt{5} )^{2} }{ {(2)}^{2} - {( \sqrt{5} )}^{2} }$

$\sf \implies p = \dfrac{4 + 5 - 2 \times 2 \times \sqrt{5}}{4- 5 }$

$\sf \implies p = \dfrac{9-4\sqrt{5}}{ - 1 }$

$\sf \implies p = 4\sqrt{5} - 9$

Now,

$\sf \implies q= \dfrac{2 + \sqrt{5} }{2 - \sqrt{5} }$

$\sf \implies q= \dfrac{(2 + \sqrt{5}) }{(2 - \sqrt{5})} \times \dfrac{(2 + \sqrt{5} )}{(2 + \sqrt{5} )}$

$\sf \implies q= \dfrac{(2 + \sqrt{5})^{2} }{ {(2)}^{2}-( \sqrt{5})^{2} }$

$\sf \implies q= \dfrac{4 + 5 + 4 \sqrt{5} }{4 - 5}$

$\sf \implies q= \dfrac{9+ 4 \sqrt{5} }{ - 1}$

$\sf \implies q= - 9 - 4 \sqrt{5}$

Therefore,

$\sf {p}^{2} - {q}^{2}$

$\sf =(p + q)(p - q)$

$\sf = \{(4 \sqrt{5} - 9) + ( - 9 - 4 \sqrt{5} ) \} \{(4 \sqrt{5} - 9) - ( - 9 - 4 \sqrt{5} ) \}$

$\sf = 18 \times 8 \sqrt{5}$

$\sf = 144 \sqrt{5}$

$\sf\implies{p}^{2} - {q}^{2} = 144\sqrt{5}$

Hence, p² - q² = 144√5

Answer:

• p² - q² = 144