Question

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

• A. 9
• B. 12
• C. 6
• D. 3

Answer 1

The correct option is D 3

Given quadratic equation is:
$p{x}^2-14x+8=0$
Let $\alpha and6\alpha$ be the roots of the given quadratic equation. Sum of the roots = $\frac{-coefficientofx}{coefficientof{x}^2}$

$\alpha +6\alpha =-(-14)/p$
$7\alpha =14/p$
$\alpha =2/p....\left(i\right)$
Product of roots = constant term /coefficient of $x^2$
$\left(\alpha \right)\left(6\alpha \right)=8/p$
$6{\alpha }^2=8/p$
Substituting $\alpha =2/p$ from (i),
$6\times (2/p{)}^2=8/p$
$24/p^2=8/p$
$3/p=1$
$p=3$
Therefore, the value of p is 3.
Option A is correct.