Find rational nos 'a' and 'b' so that
1) 3 + √7 = a + b√7
3 - √7

2) 7 + √7 - 7 - √5 = a + 7 √5b
7 - √5 7 + √5 11

3) 5 + 2√3 = a - b√3
7 + 4√3

1
In the second pro is it 7+root 5 or 7+ root7??
Oops it's 7 + root 5 sorry I'm a little absent-minded
It's ok...:)
:)

Answers

The Brainliest Answer!
2014-05-14T12:14:35+05:30
1)(3+7^1/2)/(3-7^1/2)=(3+7^1/2)(3+7^1/2)/(3-7^1/2)(3+7^1/2)
  =(3+7^1/2)^2/(3^2-7^1/2*2)    [(a+b)(a-b)=a^2-b^2]
  =(3^2+7^1/2*2+2*3*7^1/2)/9-7   [(a+b)^2=a^2+b^2+ab]
  =(9+7+6*7^1/2)/2
  =(16+6*7^1/2)/2=8+3*7^1/2=a+b*7^1/2
=>a=8 and b=3
2)[(7+5^1/2)/(7-5^1/2)]-(7-5^1/2)/7+5^1/2)
  =[(7+5^1/2)^2-(7-5^1/2)^2]/(7-5^1/2)(7+5^1/2)
  =(4*7*5^1/2)/(7^2-5^1/2*2)
  =(28*5^1/2)/49-5
  =(28*5^1/2)/44
  =7/11*5^1/2=a+7/11*5^1/2*b
=>a=0,b=1
3)(5+2*3^1/2)/(7+4*3^1/2)
  =(5+2*3^1/2)(7-4*3^1/2)/(7+4*3^1/2)(7-4*3^1/2)
  =(35-20*3^1/2+14*3^1/2-24)/(7^2-4^2*3^1/2*2)
  =(11-6*3^1/2)/(49-48)
  =11-6*3^1/2=a-b*3^1/2
=>a=11 and b=6
3 5 3
Thank you soooo much! :D
Ur wlcm ....:):)
:D