√5+√3/√5-√3=a+b√5 find a
a and b
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Answered by
6
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Anonymous:
thanks ji
Answered by
3
Heya !!!
Here is the answer.
Given :

Lets rationalize the denominator...


Now, Comparing With RHS...

a = 4 and
b√5 = √15
=> b = √15/√5 = √3
Hence, a = 4 and b=√3
Hope You Got It
Here is the answer.
Given :
Lets rationalize the denominator...
Now, Comparing With RHS...
a = 4 and
b√5 = √15
=> b = √15/√5 = √3
Hence, a = 4 and b=√3
Hope You Got It
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