If 5+√3/5-√3 = a+b√3,then find a and b
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a and b counts as 1 if there is nothing in front.
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Hi ,
LHS = ( 5 + √3 )/ ( 5 - √3 )
Rationalize the denominator ,
= [ ( 5 + √3 )(5 + √3 )]/[(5-√3)(5+√3)]
= ( 5 + √3 )²/ [( 5 )² - (√3 )² ]
= [ 5² + ( √3 )² + 2×5×√3 ]/(25 - 3 )
= ( 5 + 3 + 10√3 ) / 22
= ( 8 + 10√3 ) / 22
= 8 / 22 + (10√3 ) /22
= 4/11 + (5/11)×√3
LHS = RHS
4/11 +( 5/11 )√3 = a + b √3
Compare both sides , we get
a = 4 / 11 , b = 5 /11
I hope this helps you.
:)
LHS = ( 5 + √3 )/ ( 5 - √3 )
Rationalize the denominator ,
= [ ( 5 + √3 )(5 + √3 )]/[(5-√3)(5+√3)]
= ( 5 + √3 )²/ [( 5 )² - (√3 )² ]
= [ 5² + ( √3 )² + 2×5×√3 ]/(25 - 3 )
= ( 5 + 3 + 10√3 ) / 22
= ( 8 + 10√3 ) / 22
= 8 / 22 + (10√3 ) /22
= 4/11 + (5/11)×√3
LHS = RHS
4/11 +( 5/11 )√3 = a + b √3
Compare both sides , we get
a = 4 / 11 , b = 5 /11
I hope this helps you.
:)
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