Question

Find the value of k, for which one root of the quadratic equation kx^2+14^+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

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

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.