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

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Answer 2

Answer:

146

Step-by-step explanation:

Let the one root be α

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

So, Another root = 6α

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

Given equation : $kx^2-14x+8=0$

Sum of roots = -b/a

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

$7\alpha = \frac{14}{k}$---1

Product of roots = c/a

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

Substitute the value of α from 1 in 2

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

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

$6( \frac{196}{k})=8$

$6( \frac{196}{8})=k$

$147=k$

Hence the value of k is 146