Question

The sum of roots of x2 + px + 1 = 0 is twice the difference between them, then find p

Answer 1

for ans just look in the IMG...

Answer 2

Answer:

The value of p is $p=\sqrt{\frac{-4}{3}}$

Step-by-step explanation:

Given : The sum of roots of $x^2 + px + 1 = 0$ is twice the difference between them.

To find : The value of p ?

Solution :

Given polynomial $p(x)=x^2 + px + 1 = 0$

Here, a=1, b=p and c=1

Let the zeroes be $\alpha$ and $\beta$ respectively.

We know that,

Sum of the roots is

$\alpha+\beta=\frac{-b}{a}$

$\alpha+\beta=\frac{-p}{1}$

$\alpha+\beta=-p$ ....(1)

The sum of roots of polynomial is twice the difference between them.

i.e. $\alpha-\beta=2(\alpha+\beta)$

$\alpha-\beta=2(-p)$

$\alpha-\beta=-2p$ ....(2)

Solving (1) and (2) by adding them,

$\alpha+\beta+\alpha-\beta=-p-2p$

$2\alpha=-3p$

$\alpha=-\frac{3p}{2}$

Substitute in (1),

$-\frac{3p}{2}+\beta=-p$

$\beta=-p+\frac{3p}{2}$

$\beta=\frac{-2p+3p}{2}$

$\beta=\frac{p}{2}$

The roots are $\alpha=-\frac{3p}{2}$ and $\beta=\frac{p}{2}$

We know that,

Product of the roots is

$\alpha\beta=\frac{c}{a}$

Substitute the values,

$(-\frac{3p}{2})(\frac{p}{2})=\frac{1}{1}$

$-\frac{3p^2}{4}=1$

$p^2=\frac{-4}{3}$

$p=\sqrt{\frac{-4}{3}}$

Therefore, the value of p is $p=\sqrt{\frac{-4}{3}}$