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, equation is $p{x}^2-14x+8=0$
Let one root = $\alpha$ then other root = 6 $\alpha$
Sum of roots = - $\frac{b}{a}$
$\alpha +6\alpha =\frac{-(-14)}{p}$
7$\alpha =\frac{14}{p};$
$\alpha =\frac{14}{p\times 7}$
or $\alpha =\frac{2}{p}.....................\left(i\right)$

Products of roots = $\frac{c}{a}$
$\left(\alpha \right)\left(6\alpha \right)=\frac{8}{p}$
6 ${\alpha }^2=\frac{8}{p}........................\left(ii\right)$

Putting value of $\alpha$ from eq.(i)
6${\left(\frac{2}{p}\right)}^2=\frac{8}{p}$
$\Rightarrow 6\times \frac{4}{p^2}=\frac{8}{p}$
$\Rightarrow 24p=8p^2$
$\Rightarrow 8p^2-24p=0$
$\Rightarrow 8p(p-3)=0$
$\Rightarrow Either8p=0\Rightarrow p=0$
$\Rightarrow p-3=0$ $\Rightarrow p=3$
For p = 0, given condition is not satisfied [$\because$ leading coefficient can't be zero].
$\therefore$ p = 3