Question

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

Answer 1

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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 this quadratic equation is six times the other.

Let the roots are p and 6p.

In a quadratic equation $ax^2+bx+c=0$, the sum roots is -b/a and product of root is c/a.

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

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

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

$p=\frac{2}{k}$                      .... (1)

$p\times 6p=\frac{8}{k}$

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

Using (1), we get

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

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

$24k=8k^2$

$0=8k^2-24k$

$0=8k(k-3)$

$k=0,3$

The value of k can not be zero.

Therefore the value of k is 3.