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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!
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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imqwerty
 one year ago
Best ResponseYou've already chosen the best response.3\[(n+1)^2\times ac=n \times b^2\]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

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1@imqwerty Thank you! That really clears things up.
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