Answer :- [tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3} = a + b\sqrt3[/tex] Solving LHS :- [tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3}[/tex] [tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3} \times \dfrac{7+4\sqrt3}{7+4\sqrt3}[/tex] [tex]\implies\sf \dfrac{ (5+\sqrt3)(7+4\sqrt3)}{(7-4\sqrt3)(7+4\sqrt3)}[/tex] [tex]\implies\sf \dfrac{5( 7 + 4\sqrt3) + \sqrt3(7 + 4\sqrt3)}{ 7^2 – (4\sqrt3)^2}[/tex] [tex]\implies\sf \dfrac{35 + 20\sqrt3 + 7\sqrt3 + 4 \times 3}{49 – 48}[/tex] [tex]\implies\sf \dfrac{35 + 27\sqrt3 + 12}{1}[/tex] [tex]\implies\sf 47 + 27\sqrt3[/tex] Comparing with RHS :- [tex]\implies\sf 47 + 27\sqrt3 = a + b\sqrt3[/tex] a = 47 b = 27 Value of a = 47 and b = 27 Reply
QuestioN : (5+√3)/(7-4√3) =a+b√3 , find the value of a and b GiveN : (5+√3)/(7-4√3) =a+b√3 To FiNd : The value of a and b. ANswer : The value of a = 7 , b = – 3. SolutioN : √7 + 4√3 = √7 + 2√ ( 4 × 3 ) = √( 4 + 3 ) + 2√( 4 × 3 ) Therefore , √7 + 4√3 = √4 + √3 = 2 + √3—-( 1 ) LHS = ( 5 + √3 ) / ( √7 – 4√3 ) = ( 5 + √3 ) / ( 2 + √3 ) Rationalize the denominator = (5+√3 )(2 -√3)/(2 + √3 ) ( 2-√3 ) = [10-5√3+2√3-3]/[( 2 )²- (√3)²] = ( 7 – 3√3 )/(4 – 3 ) = 7 – 3√3 = RHS 7 – 3√3 = a + b√3 Compare both sides a = 7 , b = – 3 ∴Hence, The value of a = 7 , b = – 3. _________________________ Reply
Answer :-
[tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3} = a + b\sqrt3[/tex]
Solving LHS :-
[tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3}[/tex]
[tex]\implies\sf \dfrac{5+\sqrt3}{7-4\sqrt3} \times \dfrac{7+4\sqrt3}{7+4\sqrt3}[/tex]
[tex]\implies\sf \dfrac{ (5+\sqrt3)(7+4\sqrt3)}{(7-4\sqrt3)(7+4\sqrt3)}[/tex]
[tex]\implies\sf \dfrac{5( 7 + 4\sqrt3) + \sqrt3(7 + 4\sqrt3)}{ 7^2 – (4\sqrt3)^2}[/tex]
[tex]\implies\sf \dfrac{35 + 20\sqrt3 + 7\sqrt3 + 4 \times 3}{49 – 48}[/tex]
[tex]\implies\sf \dfrac{35 + 27\sqrt3 + 12}{1}[/tex]
[tex]\implies\sf 47 + 27\sqrt3[/tex]
Comparing with RHS :-
[tex]\implies\sf 47 + 27\sqrt3 = a + b\sqrt3[/tex]
Value of a = 47 and b = 27
QuestioN :
(5+√3)/(7-4√3) =a+b√3 , find the value of a and b
GiveN :
To FiNd :
ANswer :
The value of a = 7 , b = – 3.
SolutioN :
√7 + 4√3
= √7 + 2√ ( 4 × 3 )
= √( 4 + 3 ) + 2√( 4 × 3 )
Therefore ,
√7 + 4√3 = √4 + √3
= 2 + √3—-( 1 )
LHS = ( 5 + √3 ) / ( √7 – 4√3 )
= ( 5 + √3 ) / ( 2 + √3 )
Rationalize the denominator
= (5+√3 )(2 -√3)/(2 + √3 ) ( 2-√3 )
= [10-5√3+2√3-3]/[( 2 )²- (√3)²]
= ( 7 – 3√3 )/(4 – 3 )
= 7 – 3√3
= RHS
7 – 3√3 = a + b√3
Compare both sides
a = 7 , b = – 3
∴Hence, The value of a = 7 , b = – 3.
_________________________