Find value of a and b in 7+√5/7-√5=a+7/11√5b
Answers
Find value of a and b in 7+√5/7-√5=a+7/11√5b
LHS=\frac{7+\sqrt{5}}{7-\sqrt{5}}LHS=
7−
5
7+
5
=\frac{7+\sqrt{5}}{7-\sqrt{5}}\times\frac{7+\sqrt{5}}{7+\sqrt{5}}=
7−
5
7+
5
×
7+
5
7+
5
=\frac{(7+\sqrt{5})(7+\sqrt{5})}{(7-\sqrt{5})(7+\sqrt{5})}=
(7−
5
)(7+
5
)
(7+
5
)(7+
5
)
=\frac{7^2+(\sqrt{5})^2+14\sqrt{5}}{7^2-(\sqrt{5})^2}=
7
2
−(
5
)
2
7
2
+(
5
)
2
+14
5
=\frac{49+5+14\sqrt{5}}{49-5}=
49−5
49+5+14
5
=\frac{54+14\sqrt{5}}{44}=
44
54+14
5
=\frac{54}{44}+\frac{14\sqrt{5}}{44}=
44
54
+
44
14
5
=\frac{27}{22}+\frac{7\sqrt{5}}{22}=
22
27
+
22
7
5
RHS=a+\frac{7}{11}\sqrt{5}bRHS=a+
11
7
5
b
\implies a=\frac{27}{22}⟹a=
22
27
\implies \frac{7}{11}\sqrt{5}b=\frac{7\sqrt{5}}{22}⟹
11
7
5
b=
22
7
5
b=\frac{1}{2}b=
2
1
Therefore, a=\frac{27}{22}\:\:and\:\:b=\frac{1}{2}a=
22
27
andb=
2
1
Answer:
of a and b in 7+√5/7-of a and b in 7+√5/7-√5=a+7/11√of a and b in 7+√5/7-√5=a+7/11√5bof a and b in 7+√5/7-√5=a+7/11√5b5b√5of a and b in 7+√5/7-√5=a+7/11√5b=a+7/11√5b