Question

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

Answer 1

Let the roots be a and b & a = 6b.
Therefore the sum of roots $=a+b=7b=-\frac{(-14)}{k}$
$\Rightarrow b=\frac{2}{k}.....\left(i\right)$
product of roots $=ab=6b^2=\frac{8}{k}$
$\Rightarrow b=\sqrt{\frac{8}{6k}}=\sqrt{\frac{4}{3k}}....\left(ii\right)$
Equating (i) and (ii),
$\frac{2}{k}=sqrt\left(\frac{4}{3k}\right)\frac{4}{k^2}=\frac{4}{3k}\Rightarrow k^2=3kork=3.$