Question

Find the value of p and q if $\frac{7+\sqrt{5} }{7-\sqrt{5} } - \frac{7-\sqrt{5}}{7+\sqrt{5} } = p-7\sqrt{5} q$

Answer 1

$</p><p> \sf \rightarrow \quad \frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7 \sqrt{5}q \\ \\ \sf \rightarrow \quad \frac{ {(7 + \sqrt{5} )}^{2} - {(7 - \sqrt{5}) }^{2} }{(7 - \sqrt{5} )(7 + \sqrt{5} )} = p - 7 \sqrt{5}q \\ \\ \sf \rightarrow \quad \frac{(7 + \sqrt{5} + 7 - \sqrt{5})(7 + \sqrt{5} - 7 + \sqrt{5} )}{49 - 5} = p - 7 \sqrt{5} q \\ \\ \sf \rightarrow \quad \frac{14(2 \sqrt{5}) }{44} = p - 7 \sqrt{5} q \\ \\ \sf \rightarrow \quad \frac{7 \sqrt{5} }{11} = p - 7 \sqrt{5} q \\ \\ \sf \rightarrow \quad \frac{14 \sqrt{5} }{11} - \frac{7 \sqrt{5} }{11} = p - 7 \sqrt{5} q \\ \\ \sf \rightarrow \quad \frac{14 \sqrt{5} }{11} - 7 \sqrt{5} \cdot \frac{1}{11} = p - 7 \sqrt{5} q \\ \\ \\ </p><p>$

$</p><p></p><p>\sf On \: comparing, \\ \\</p><p>\sf p = \frac{ 14\sqrt{5} }{11} \quad and \quad q = \frac{1}{11}</p><p></p><p>$

Answer 2

Question :–

• Find the value of p and q if $\\ { \bold{ \dfrac{7+\sqrt{5} }{7-\sqrt{5} } - \dfrac{7-\sqrt{5}}{7+\sqrt{5} } = p-7\sqrt{5} q }}$

ANSWER :–

$\\ \to \: \: \: { \bold{ p = 0 \: \: , \: \: q = - \frac{1}{11}}}$

EXPLANATION :–

GIVEN :–

• A relation ${ \bold{ \dfrac{7+\sqrt{5} }{7-\sqrt{5} } - \dfrac{7-\sqrt{5}}{7+\sqrt{5} } = p-7\sqrt{5} q }}$

TO FIND :–

• Value of p and q.

SOLUTION :–

$\\ \implies { \bold{ \dfrac{7+\sqrt{5} }{7-\sqrt{5} } - \dfrac{7-\sqrt{5}}{7+\sqrt{5} } = p-7\sqrt{5} q }}$

• We should write this as –

$\\ \implies { \bold{ \dfrac{(7+\sqrt{5})^{2} - {(7 - \sqrt{5} )}^{2} } {(7-\sqrt{5})(7 + \sqrt{5} ) } = p-7\sqrt{5} q }}$

• We know that –

$\\ \: \: \: \: \: . \: \: \: { \bold{ (a + b)^{2} = {a}^{2} + {b}^{2} + 2ac }}$

$\\ \: \: \: \: \: . \: \: \: { \bold{ (a - b)^{2} = {a}^{2} + {b}^{2} - 2ac }}$

$\\ \: \: \: \: \: . \: \: \: { \bold{ {a}^{2} - {b}^{2} = (a + b)(a - b) }}$

• So that –

$\\ \implies { \bold{ \dfrac{49 + 5 + 14 \sqrt{5} - {(49 + 5 - 14 \sqrt{5} )} } {[ 7^{2} - ( \sqrt{5}) ^{2}] } = p-7\sqrt{5} q }}$

$\\ \implies { \bold{ \dfrac{54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} } {49 - 5 } = p-7\sqrt{5} q }}$

$\\ \implies { \bold{ \dfrac{28 \sqrt{5} } {44 } = p-7\sqrt{5} q }}$

$\\ \implies { \bold{ \dfrac{7 \sqrt{5} } {11 } = p-7\sqrt{5} q }}$

• Now let's compare –

$\\ \implies { \boxed { \bold{ p = 0 \: \: , \: \: q = - \frac{1}{11}}}}$