If 1/(2√3+√7) =a√3 +b√7, then a and b are
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Answer:
a = 2/5 , b = -1/5
Solution:
- Given : 1/(2√3 + √7) = a√3 + b√7
- To find : a , b = ?
We have ;
1/(2√3 + √7) = a√3 + b√7
Thus,
a√3 + b√7 = 1/(2√3 + √7)
Now,
Rationalising the denominator of the term in RHS , we have ;
a√3 + b√7
= 1/(2√3 + √7)
= (2√3 - √7)/(2√3 + √7)(2√3 - √7)
= (2√3 - √7)/[ (2√3)² - (√7)² ]
= (2√3 - √7)/(12 - 7)
= (2√3 - √7)/5
= 2√3/5 - √7/5
= (2/5)•√3 + (-1/5)•√7
Thus,
a√3 + b√7 = (2/5)•√3 + (-1/5)•√7
On comparing like terms both the sides ,
We have ; a = 2/5 and b = -1/5
Hence,
a = 2/5
b = -1/5
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