**the roots of \(ax^{2}\)+bx+c=0 ate in ratio m:n**

let the roots be mk and nk respectively

=>a \((mk)^{2}\)+bmk+c=0

=>a \(m^{2 } k^{2}\)+bmk+c=0.....(1)

Similarly,

a \(n^{2}\)\( k^{2}\)+bnk+c=0......(2)

(1)-(2)

=>a\( k^{2}\)(\(m^{2 }\)-\(n^{2}\))+bk(m-n)=0

=>a\( k^{2}\)(m+n)(m-n)+bk(m-n)=0

=>k(m-n)(ak(m+n)+b)=0

=>ak(m+n)+b=0

=>ak(m+n)=-b

=>k=-\(\frac {b}{a(m+n)}\)

therefore,

the roots are mk=-\(\frac{mb}{a(m+n)}\)

(or)

nk=-\(\frac{nb}{a(m+n)} \)

## Comments

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TopNewestAfter letting one root be \( mk \) and other be \( nk \).

You could have directly used sum of roots = \( \dfrac{-b}{a} \)

\( \therefore k = \dfrac{-b}{a(m+n)} \)

Roots are,

\( \dfrac{-mb}{a(m+n)} \) and \( \dfrac{-nb}{a(m+n)} \) – Vighnesh Shenoy · 10 months, 1 week ago

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