Question

If p=2-root5/2+root5 and q =2+root 5/2-root5 find the values of p+q 2ndp-q

Answer 1

$Given \: p = \frac{(2-\sqrt{5})}{(2+\sqrt{5})} \\and \: q = \frac{(2+\sqrt{5})}{(2-\sqrt{5})}$

$\red{i ) p + q } \\ = \frac{(2-\sqrt{5})}{(2+\sqrt{5})} + \frac{(2+\sqrt{5})}{(2-\sqrt{5})}\\= \frac{ (2-\sqrt{5})^{2} + (2+\sqrt{5})^{2}}{(2+\sqrt{5})(2-\sqrt{5})} \\= \frac{2[2^{2} + (\sqrt{5})^{2}]}{2^{2} - (\sqrt{5})^{2} }$

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By Algebraic Identities:

1 . (a+b)² + (a-b)² = 2(a²+b²)

2. (a+b)(a-b) = a² - b²

3. (a+b)² - (a-b)² = 4ab

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$= \frac{ 2( 4 +5 )}{4-5} \\= \frac{ 18}{-1} \\= - 18$

$\red{ii ) p - q } \\= \frac{(2-\sqrt{5})}{(2+\sqrt{5})} - \frac{(2+\sqrt{5})}{(2-\sqrt{5})}\\= \frac{ (2-\sqrt{5})^{2} - (2+\sqrt{5})^{2}}{(2+\sqrt{5})(2-\sqrt{5})} \\= \frac{-4 \times 2 \times \sqrt{5} }{2^{2} - (\sqrt{5})^{2} }$

$= \frac{-8\sqrt{5} }{(4-5)} \\= 8\sqrt{5} \:--(2)$

Therefore.,

$\red{i )p + q } \green {= -18 }$

$\red{ii ) p - q } \green {= 8\sqrt{5} }$

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