Question

Find the value of $p$, for which one root of the quadratic equation $p{x}^2-14x+8=0$ is $6$ times the other

Answer 1

Given quadratic equation,

$p{x}^2-14x+8=0$

here,

let one root of the given equation be $m.$

other root=$6m.$

Now,sum of the roots=$\frac{-b}{a}$

$=>m+6m=-\frac{-14}{p}$

solving $p=\frac{2}{m}---------\left(i\right)$

proctuct of roots$=\frac{c}{a}$

Now,

$=>m\times 6m=\frac{8}{p}$

$p=\frac{4}{3m^2}---------\left(ii\right)$

Here, put the value of $p=\frac{2}{m}$ from (i),we get,

$=>\frac{2}{m}=\frac{4}{3m^2}$

$=>3m=2$

$=>m=\frac{2}{3}----------\left(iii\right)$

Here,equating equation $\left(i\right)$ and $\left(iii\right)$ we get,

$=>p=\frac{2}{m}$

$=>m=\frac{2}{p}$

$=>\frac{2}{p}=\frac{2}{3}$

$=>p=3$