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## Jadedry one year ago Given that the roots of the equation ( ax^2 +bx +c = 0 ) are beta and (n * beta) show that )((n+1)^2 * ac) =n*b^2 I've figured out that: beta + (n * beta) = -b/a n * beta^2 = c/a (beta + (n * beta) / n*beta^2) = b/c) Unfortunately, that's as far as I got, I'm not sure about to continue. Could someone please provide some hints? Thanks in advance!

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1. imqwerty

$(n+1)^2\times ac=n \times b^2$divide both sides by a^2$(n+1)^2 \times \frac{ c }{ a }=n \times \left( \frac{ b }{ a } \right)^2$ we know that $\frac{ c }{a }=n*\beta^2$ and $\frac{ -b }{ a }=n \beta+\beta$$\frac{ b }{ a }=-(n \beta+\beta)$now just put these two values in the equation u get- $(n+1)^2 \times n \beta^2 =n \times [-(n \beta+\beta)]^2$ $(n+1)^2 \times n \beta^2 = n \times [-\beta(n+1)]^2$$(n+1)^2 \times n \beta^2 =n \times \beta^2(n+1)^2$LHS=RHS hence proved

2. Jadedry

@imqwerty Thank you! That really clears things up.

3. imqwerty

:) no prblm

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