If (7 + √5)/(7 - √5) = a + 7√5b , then a , b = ?
Answers
We have ;
- (7 + √5)/(7 - √5) = a + 7√5b
Rationalising the denominator of the term in LHS , we get ;
= (7 + √5)²/(7 - √5)(7 + √5) = a + 7√5b
= [7² + (√5)² + 2•7•√5]/[7² - (√5)²] = a + 7√5b
= (49 + 5 + 14√5)/(49 - 5) = a + 7√5b
= (54 + 14√5)/44 = a + 7√5b
= 54/44 + 14√5/44 = a + 7√5b
= 27/22 + 7√5/22 = a + 7√5b
= 27/22 + 7√5•(1/22) = a + 7√5b
Now ,
Comparing both the sides , we get ;
a = 27/22 , b = 1/22
______________________________
Step-by-step explanation:
a=
22
27
andb=
2
1
Step-by-step explanation:
Given: \frac{7+\sqrt{5}}{7-\sqrt{5}}=a+\frac{5}{11}\sqrt{5}b
7−
5
7+
5
=a+
11
5
5
b
To find: value of a and b
First we rationalize the denominator of LHS and them compare the answer with RHS to find value of a & b
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