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

#Algebra

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Comments

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 $\dfrac{-nb}{a(m+n)}$

A Former Brilliant Member - 5 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$