Question

indices: a²+(p+1/p)a+1

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Answer 1

Answer:

a²+(p+1/p)a+1=0 ➜ a²+a(p²/p+1/p)+1 ->

By Shreedharacharya's rule,

a= {-(p+1/p)±√[(p²+1/p²+2p×1/p)-4]}/2

→ a= {-(p+1/p)± √[(p²+1/p²+2-4)]}/2

→ a= {-(p+1/p)±√[(p²+1/p²-2]}/2

→ a= [-(p+1/p)± √(p-1/p)²]/2

→ a= [-(p+1/p)±(p-1/p)]/2

= [(-p-1/p)±(p-1/p)]/2

⇒a=

= [(-p-1/p)+(p-1/p)]/2, [(-p-1/p) ⇒a=

(p-1/p)]/2

→ a= (-p-1/p+p-1/p)/2, (-p-1/p- p+1/p)/2

a= (-1/p-1/p)/2, (-p-p)/2

→ a= (-2/p)/2, (-2p)/2

→ a= (-1/p), (-p)