Question

if a²-5a+1=0 and a≠0, find: (i) a+1/a (ii) a²+1/a²​

Answer 1

Answer:

a2−5a−1=0

=>a2−1=5a

=>a−a1=5

Squaring both sides, we get

=>a2+a21−2a×a1=25

=>a2+a21=25+2

=>a2+a21=27

Now,

(a+a1)2=a2+a21+2a×a1

                       =a2+a21+2

                       =27+2

                       =29

Thus, a+a1=±29

Answer 2

Answer:

a^2+5a_1=0

a^2_1=5a

a-1/a=5

(a+1/a)^2=(a-1/a)^2+4a×1/a

=5^2+4=29

a+1/a=root over 29

a^2_1/a^2=(a+1/a)(a-1/a)

=root over29×5

=5root over29