Question

find the value of k for which one root of the quadratic equation kx2 - 14 x + 8 = 0 is 6 times the other

Answer 1

Hope this answer may help you

Answer 2

Answer:

The value of k is 3.

Step-by-step explanation:

The given equation is

$kx^2-14x+8=0$

It is given that one root of the quadratic equation is 6 times the other.

Let p and 6p are two roots.

If α and β are two roots of the equation $ax^2+bx+c=0$, then

$\alpha +\beta =\frac{-b}{a}$

$\alpha\beta =\frac{c}{a}$

$p+6p=-\frac{-14}{k}$

$7p=\frac{14}{k}$

$p=\frac{2}{k}$

$6p^2=\frac{8}{k}$

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

$6(\frac{4}{k^2})=\frac{8}{k}$

$\frac{24}{k^2}=\frac{8}{k}$

$24k=8k^2$

$8k^2-24k=0$

$8k(k-3)=0$

$k=0,3$

The value of k can not be 0.

Therefore the value of k is 3.