Question

If 7+3 root 5/3+ root 5 - 7-3 root 5/3- root 5=p+q root 5 Find value of p and q​

Answer 1

Step-by-step explanation:

7+3√5/3 +√5 - 7-3√5/3 - √5 = p+q√5

LHS

= 7+ √5 +√5 - 7 - √5 - √5

= (7-7) + ( 2√5 - 2√5)

= 0 + 0

= 0

Therefore RHS =>

p+q√5= 0

=> p : q = - √5 : 1

thus infinite number of solutions can be found

which ratios are -√5 : 1.

Answer 2

$\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = p + q \sqrt{5} \\ \\ \frac{(7 + 3 \sqrt{5} )(3 - \sqrt{5}) - (7 - 3 \sqrt{5} )(3 + \sqrt{5} ) }{9 - 5} = p + q \sqrt{5} \\ \\ \frac{21 - 7 \sqrt{5} + 9 \sqrt{5} - 15 - 21 - 7 \sqrt{5} + 9 \sqrt{5} + 15 }{4} = p + q \sqrt{5} \\ \\ \frac{4 \sqrt{5} }{4} = p + q \sqrt{5} \\ \\ \sqrt{5} = p + q \sqrt{5} \\ \\ comparing \: both \: sides \: we \: get \\ p = 0 \: \: \: \: \: \: \: q = 1$