Question

Find the condition so that zeros of ax² + bx + c are in the ratio 3:4

Answer 1

$let \: the \: roots \: of \: the \: equation \\ \\ a {x}^{2} + bx + c = 0 \\ \\ be \: \: \alpha \: \: and \: \: \beta \\ \\ wkt \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \beta = \frac{c}{a} \\ \\ given \\ \\ \frac{ \alpha }{ \beta } = \frac{3}{4} \\ \\ 4 \alpha = 3 \beta \\ \\ \alpha = \frac{3 \beta }{4} \\ \\ \frac{3 \beta }{4} + \beta = - \frac{b}{a} \\ \\ \beta = - \frac{4b}{7a} \\ \\ ( \frac{3 \beta }{4} )( \beta ) = \frac{c}{a} \\ \\ { \beta }^{2} = \frac{4c}{3a} \\ \\ \frac{16 {b}^{2} }{49 {a}^{2} } = \frac{4c}{3a} \\ \\ 48 {b}^{2} = 196ac \\ \\ 12 {b}^{2} = 49ac$

hope it helps.

Answer 2

Step-by-step explanation:

$\begin{lgathered}let \: the \: roots \: of \: the \: equation \\ \\ a {x}^{2} + bx + c = 0 \\ \\ be \: \: \alpha \: \: and \: \: \beta \\ \\ wkt \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \beta = \frac{c}{a} \\ \\ given \\ \\ \frac{ \alpha }{ \beta } = \frac{3}{4} \\ \\ 4 \alpha = 3 \beta \\ \\ \alpha = \frac{3 \beta }{4} \\ \\ \frac{3 \beta }{4} + \beta = - \frac{b}{a} \\ \\ \beta = - \frac{4b}{7a} \\ \\ ( \frac{3 \beta }{4} )( \beta ) = \frac{c}{a} \\ \\ { \beta }^{2} = \frac{4c}{3a} \\ \\ \frac{16 {b}^{2} }{49 {a}^{2} } = \frac{4c}{3a} \\ \\ 48 {b}^{2} = 196ac \\ \\ 12 {b}^{2} = 49ac \\ \\\end{lgathered}$