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22 votes
22 votes
Given that the roots of the equation a
x^(2)+b
x+c=0 are
\beta and n
\beta, show that (n+1)²ac=nb²

User Armonge
by
2.6k points

1 Answer

15 votes
15 votes

Answer:

Explanation:

sum of roots β+nβ=-b/a

(n+1)β=-b/a

squaring

(n+1)²β²=b²/a²

product of roots β×nβ=c/a

nβ²=c/a

β²=c/na

∴(n+1)²c/na=b²/a²

multiply by na²

(n+1)²ac=nb²

User Ando Saabas
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