Question

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!

Answer 1

\[(n+1)^2\times ac=n \times b^2\]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

Answer 2

@imqwerty

Thank you! That really clears things up.

Answer 3

:) no prblm