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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply