Question

Find the value of k for one root of the quadratic equation kx²-14x+8=0 is six times the other?​

Answer 1

let one root be n then other root will be 6n
now.,
sum of roots
$n + 6n = \frac{14}{k} \\ n = \frac{2}{k}$
product of roots
$n(6n) = \frac{8}{k} \\ 6 {n}^{2} = \frac{8}{k} \\ from \: (1) \\ 6( \frac{2}{k} ) {}^{2} = \frac{8}{k} \\ \frac{6}{ {k}^{2} } = \frac{2}{k} \\ \frac{3}{k {}^{2} } = \frac{1}{k} \\ now \: k \: can \: not \: be \: equal \: to \: zero \\ \frac{3}{k} = 1 \\ k = 3$

so k =3 (k shouldn't be equal to zero)

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