Question

factorise: a2 + (p + 1/p) a + 1. ( step by step explanation)​

Answer 1

Answer:

Step-by-step explanation:

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

→a= [(-p-1/p)±(p-1/p)]/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- p+1/p)/2

→a= (-1/p-1/p)/2, (-p-p)/2

→a= (-2/p)/2, (-2p)/2

→a= (-1/p), (-p)

Hope this helps.