Question

Find the value of k, for which one root of the quadratic equation kx2-14x+8=0 is six times the other.

Answer 1

it would be 7 as putting value as 6x into the quadratic equ. and solving by quadratic formua we would get the  answer to be 7. 

Answer 2

Answer:

147

Step-by-step explanation:

General equation : $ax^2+bx+c=0$

We are given an equation $kx^2-14x+8=0$

Let the one root of given equation be α

We are given that another root is 6 times of one root

So, Another root = 6α

Product of roots = c/a

So, $6\alpha^2 = \frac{8}{k}----1$

Sum of roots = -b/a

So, $\alpha + 6\alpha = \frac{14}{k}$

$7\alpha = \frac{14}{k}---2$

Substitute the value of α from 2 in 1

$6( \frac{14}{7k})^2= \frac{8}{k}\\\\6( \frac{196}{k^2})= \frac{8}{k}\\\\6( \frac{196}{k})=8\\\\6( \frac{196}{8})=k\\\\147=k$

So, the value of k is 147