Question

If one root of $p{x}^2-14x+8=0$ is $6$ times the other, then $p=$

• A. $3$
• B. $-3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

$\Rightarrow$ The given quadratic equation is $p{x}^2-14x+8=0$

$\Rightarrow$ Let the root be $\alpha$ then other root will be $6\alpha$

$\Rightarrow$ Sum of the roots $=\frac{-b}{a}$

$\Rightarrow$ $\alpha +6\alpha =\frac{-(-14)}{p}$

$\Rightarrow$ $7\alpha =\frac{14}{p}$

$\therefore$ $\alpha =\frac{2}{p}$ -------- ( 1 )

$\Rightarrow$ Product of the roots $=\frac{c}{a}$

$\Rightarrow$ $\alpha .6\alpha =\frac{8}{p}$

$\Rightarrow$ $6{\alpha }^2=\frac{8}{p}$

$\Rightarrow$ $6{\left(\frac{2}{p}\right)}^2=\frac{8}{p}$ [ From ( 1 ) ]

$\Rightarrow$ $6\times \frac{2}{p^2}=\frac{8}{p}$

$\therefore$ $p=3$