Question

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

Answer 1

Given
$p{x}^2-14x+8=0$
Let $\alpha ,\beta$ are the roots of the equation
$\beta =6\alpha$
we know that
sum of the roots $=\frac{-b}{a}$
product of the roots $=\frac{c}{a}$
here a = p, b = -14, c = 8
$\Rightarrow \alpha +\beta =\frac{14}{p}\alpha ,\beta =\frac{8}{p}\Rightarrow \alpha +6\alpha =\frac{14}{p}\Rightarrow 7\alpha =\frac{14}{p}\Rightarrow \alpha =\frac{2}{p}\alpha \beta =\frac{8}{p}\Rightarrow \alpha \left(6\alpha \right)=\frac{8}{p}\Rightarrow 6{\alpha }^2=\frac{8}{p}\Rightarrow 3{\alpha }^2=\frac{4}{p}\Rightarrow 3{\left(\frac{2}{p}\right)}^2=\frac{4}{p}\Rightarrow 3\left(\frac{4}{p^2}\right)=\frac{4}{p}\Rightarrow p=3$