determine 5+root 3/7-4root3=a+3root b
Answers
Step-by-step explanation:
Answer:
The value of a=\frac{1}{2}a=
2
1
and b=9b=9
Step-by-step explanation:
Given : Expression \frac{5+\sqrt3}{7-4\sqrt3}=94a+3\sqrt{3}b
7−4
3
5+
3
=94a+3
3
b
To find : The value of a and b?
Solution :
Solving the LHS of the expression,
\frac{5+\sqrt3}{7-4\sqrt3}
7−4
3
5+
3
Rationalize,
=\frac{5+\sqrt3}{7-4\sqrt3}\times \frac{7+4\sqrt3}{7+4\sqrt3}=
7−4
3
5+
3
×
7+4
3
7+4
3
=\frac{(5+\sqrt3)(7+4\sqrt3)}{(7-4\sqrt3)(7+4\sqrt3)}=
(7−4
3
)(7+4
3
)
(5+
3
)(7+4
3
)
=\frac{35+20\sqrt3+7\sqrt3+4\times 3}{7^2-(4\sqrt3)^2}=
7
2
−(4
3
)
2
35+20
3
+7
3
+4×3
=\frac{47+27\sqrt3}{49-48}=
49−48
47+27
3
=\frac{47+27\sqrt3}{1}=
1
47+27
3
=47+27\sqrt3=47+27
3
On comparing with RHS,
47+27\sqrt3=94a+3\sqrt{3}b47+27
3
=94a+3
3
b
94a=4794a=47
a=\frac{47}{94}a=
94
47
a=\frac{1}{2}a=
2
1
and 3b=273b=27
b=\frac{27}{3}b=
3
27
b=9b=9
Therefore, The value of a=\frac{1}{2}a=
2
1
and b=