Question

if one zeroe of polynomial 3x^2+(1+4k)x+k^2+5 is one third of the other find k

Answer 1

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Answer 2

Answer:

The value of k is $\frac{79}{8}$.

Step-by-step explanation:

The given polynomial is

$3x^2+(1+4k)x+k^2+5$

It is given that the one zero of polynomial is one third of the other .

Let the zeroes of the given polynomial are x and x/3.

If a polynomial is defined as $ax^2+bx+c=0$, then

1. The sum of zeroes is -b/a.

2. The product of zeroes is c/a.

$x+\frac{x}{3}=-\frac{1+4k}{3}$

$\frac{4x}{3}=-\frac{1+4k}{3}$

$4x=-(1+4k)$

$x=-\frac{(1+4k)}{4}$                 .... (1)

$x\times \frac{x}{3}=-\frac{k^2+5}{3}$

$x^2=k^2+5$

Using (1), we get

$(-\frac{(1+4k)}{4})^2=k^2+5$

$\frac{1}{16}+\frac{x}{2}+k^2=k^2+5$

$\frac{x}{2}=5-\frac{1}{16}$

$\frac{x}{2}=\frac{79}{16}$

Multiply both sides by 2.

$\frac{x}{2}=\frac{79}{8}$

Therefore the value of k is $\frac{79}{8}$.