Factorise: a^2+(p+1/p)a+1
Answers
Answered by
1
Answer:
Step-by-step explanation:
-( p+1/p) and 1
Answered by
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)
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