Given that the roots of the equation ax^(2)+bx+c=0 are \beta and n\beta, show that (n+1)²ac=nb²

Question

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

Answer 1

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