Question

Find the value of k in order that one zero of polynomial 3x2+(1+4k)x+k2+5 maybe one-third of the other

Answer 1

Answer:

The value of k is 79/8.

Step-by-step explanation:

Since, we know that, if m and n are the zeroes of a quadratic equation,

Then,

$m+n=-\frac{\text{Coefficient of }x}{\text{Coefficient of }x^2}$

$m.n=\frac{\text{Constant term}}{\text{Coefficient of }x^2}$

Here, the given quadratic equation is,

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

$m+n=-\frac{1+4k}{3}$-----(1),

$m.n=\frac{k^2+5}{3}$ ----(2),

According to the question,

$m=\frac{n}{3}$ ------(3),

From equation (1) and (3),

We get,

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

$\implies n=-\frac{1+4k}{4}$ -----(4),

From equation (2) and (3),

We get,

$\frac{n^2}{3}=\frac{k^2+45}{3}$

$\implies n=\sqrt{k^2+4}$ ----(5),

From equation (4) and (5),

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

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

By squaring,

$1+16k^2+8k=16k^2+80$

$1+8k=80$

$8k=79\implies k=\frac{79}{8}$

Answer 2

Answer:

Step-by-step explanation: