Question

Question #2be34

Answer 1

This equation has 1 integer root : `x_1=-2` and 2 irrational roots:

`x_2=(-1-sqrt(5))/2`

`x_3=(-1+sqrt(5))/2`

Explanation:

There is a theorem, that says:

"If an integer is a root of a polynomial with integer coefficients, then it is a divisor of the free term of that polynomial."

So if you know, that the polynomial has an integer root you can only look for it among divisors of free term (in this case `-2`)

This is how you find the integer root. To calculate other roots you have to divide the polynomial by `x+2`.

After dividing you find out thast the initial polynomial can be written as:

`P(x)=(x-2)(x^2+x-1)`

So yo find the remaining roots you have to solve the equation

`x^2+x-1=0`

`Delta=1-4*1*(-1)=5`

`sqrt(Delta)=sqrt(5)`

`x_2=(-1-sqrt(5))/2`

`x_3=(-1+sqrt(5))/2`