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roots in m:n

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

Note by Madhav Rockzz
7 months, 3 weeks ago

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After 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 · 7 months, 3 weeks ago

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