Question

If p=root 7- root5/root7+root5 and q=root7+root5/root7-root5 Find the value of p2-q2

Answer 1

Hey friend,
Here is the answer you were looking for:
$p = \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \\ \\ on \: rationalizing \: we \: get \\ \\ = \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \times \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \\ \\ using \: the \: idenities \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{7} )}^{2} + {( \sqrt{5} )}^{2} - 2 \times \sqrt{7} \times \sqrt{5} }{ {( \sqrt{7} )}^{2} - {( \sqrt{5}) }^{2} } \\ \\ = \frac{7 + 5 - 2 \sqrt{35} }{7 - 5} \\ \\ = \frac{12 - 2 \sqrt{35} }{2} \\ \\ = 6 - \sqrt{35} \\ \\ q = \frac{ \sqrt{7} + \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \times \frac{ \sqrt{7} + \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \\ \\ using \: the \: idenities \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{7}) }^{2} + {( \sqrt{5}) }^{2} + 2 \times \sqrt{7} \times \sqrt{5} }{ {( \sqrt{7} })^{2} - {( \sqrt{5} )}^{2} } \\ \\ = \frac{7 + 5 + 2 \sqrt{35} }{7 - 5} \\ \\ = \frac{12 + 2 \sqrt{35} }{2} \\ \\ = 6+ \sqrt{35} \\ \\ {p}^{2} - {q}^{2} \\ \\ = {(6 - \sqrt{35} )}^{2} - {(6 + \sqrt{35} )}^{2} \\ \\ = ( {(6)}^{2} + {( \sqrt{35} )}^{2} - 2 \times 6 \times \sqrt{35} ) - ( {(6)}^{2} + {( \sqrt{35}) }^{2} + 2 \times 6 \times \sqrt{35} ) \\ \\ = (36 + 35 - 12 \sqrt{35} ) - (36 + 35 + 12 \sqrt{35} ) \\ \\ = (71 - 12 \sqrt{35} ) - (71 + 12 \sqrt{35} ) \\ \\ = 71 - 12 \sqrt{35} - 71 - 12 \sqrt{35} \\ \\ = - 12 \sqrt{35} - 12 \sqrt{35} \\ \\ = - 24 \sqrt{35}$

Hope this helps!!!!

@Mahak24

Thanks....
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