Question

If one root of the quadratic equation kx2-14x+8=0 is 6 times the other, then find the value of k.

Answer 1

heya ....

ur ans is in this attachment.....

Answer 2

Answer:

The value of k is 3.

Step-by-step explanation:

The given equation is

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

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

$\alpha+\beta=\frac{-b}{a}$          .... (1)

$\alpha \beta=\frac{c}{a}$            ..... (2)

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

Let one roots be a and other is 6a.

Using (1) and (2) we get

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

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

$k=\frac{14}{7a}=\frac{2}{a}$              .... (3)

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

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

$k=\frac{8}{6a^2}=\frac{4}{3a^2}$         ....(4)

Equating (3) and (4).

$\frac{2}{a}=\frac{4}{3a^2}$

$6a^2=4a$

$6a^2-4a=0$

$2a(3a-2)=0$

$a=0,\frac{2}{3}$

At a=0, k is undefined.

At $a=\frac{2}{3}$,

$k=\frac{2}{\frac{2}{3}}=3$

Therefore the value of k is 3.