Question

find the value of a and b in the following if: 5+√5/9-4√5= a + b√5​

Answer 1

Answer:= 9^3 +4√5^3+3ab(a+b) +9^3-4√5^3-3ab(a-b)

Step-by-step explanation:If a = 1 /9-4√5 and b = 1 /9+4√5 find the value of the following: (i) a 2 + b 2 (ii) a3 + b3

given, a=9+4√5 & b=1/9+4√5

therefore ..b=1×9-4√5/9+4√5×9-4√5

b=9-4√5/81-80

b=9-4√5

now....

a^3+b^3= (9+4√5)^3+(9-4√5)^3

= 9^3 +4√5^3+3ab(a+b) +9^3-4√5^3-3ab(a-b)

hope it helps !!

Answer 2

Answer:

a=65 b=29

Step-by-step explanation:

$\frac{5+\sqrt{5} }{9-4\sqrt{5} }$  =a+b$\sqrt{5}$

multiplying numerator and denominator by (9+4$\sqrt{5}$)

$\frac{5+\sqrt{5} }{9-4\sqrt{5} } *\frac{9+4\sqrt{5} }{9+4\sqrt{5} }\\\frac{45+20\sqrt{5}+9\sqrt{5}+20 }{9^{2} -(4\sqrt{5}) ^{2} \\} }\\\frac{65+29\sqrt{5} }{81-80} \\\frac{65+29\sqrt{5} }{1} =a+b\sqrt{5} \\a=65, b=29$