Question

Solution please.......​

Answer 1

Answer:

p = 0

q = - 1/11

Explanation:

Given,

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

Consider LHS,

$\tt \dfrac{7 +\sqrt{5}}{7-\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}$

Rationalising the denominator;

$\tt \dfrac{7 +\sqrt{5}}{7-\sqrt{5}}*\dfrac{7 +\sqrt{5}}{7+\sqrt{5}}-\dfrac{7-\sqrt{5}}{7+\sqrt{5}}*\dfrac{7 - \sqrt{5}}{7-\sqrt{5}}$

$\tt \longrightarrow \dfrac{(7 +\sqrt{5})^2}{(7)^2-(\sqrt{5})^2}-\dfrac{(7 - \sqrt{5})^2}{(7)^2-(\sqrt{5})^2}$

[ ∵ (a + b)(a - b) = a² - b² ]

$\tt \longrightarrow \dfrac{49 +14\sqrt{5}+5}{49-5}-\dfrac{49 - 14\sqrt{5}+5}{49-5}$

$\tt \longrightarrow \dfrac{54 +14\sqrt{5}}{44}-\dfrac{ 54- 14\sqrt{5}}{44}$

$\tt \longrightarrow \dfrac{54 +14\sqrt{5}-54+14\sqrt{5}}{44}$

$\tt \longrightarrow \dfrac{28\sqrt{5}}{44}$

$\boxed{\bf \longrightarrow \dfrac{7\sqrt{5}}{11}}$

Comparing LHS with RHS, we get -

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

Therefore, the values of p and q are :

• p = 0
• q = $\tt \dfrac{-1}{11}$

Answer 2

$q = \frac{ - 1}{11}$