Question

Find value of a and b in 7+√5/7-√5=a+7/11√5b
​

Answer 1

$\huge\mathfrak\red{question:)}$

Find value of a and b in 7+√5/7-√5=a+7/11√5b

$\huge\mathfrak\red{answer:)}$

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 2

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