2 + 5 , 1 are roots of the cubic equation given by

Question

Aicharya · Accepted Answer

We know that, in a cubic equation with rational coefficients has a irrational root $α + \sqrt{β}$ , where $α$ and $β$ are rational and $β$ is not a perfect square, then it has also a conjugate root $α - \sqrt{β}$ .

Hence, $1, 2 + \sqrt{5}, 2 - \sqrt{5}$ are the roots of cubic equation.

We know that, if $α, β, γ$ are the roots of cubic equation, then its equation is $x^{3} - (α + β + γ) x^{2} + (α β + α γ + β γ) x - (α β γ)$ .

Let, $α = 1, β = 2 + \sqrt{5}, γ = 2 - \sqrt{5}$

$α + β + γ = 1 + 2 + \sqrt{5} + 2 - \sqrt{5} = 5$ , $α β + α γ + β γ = = 1 (2 + \sqrt{5}) + (2 + \sqrt{5}) (2 - \sqrt{5}) + 1 (2 - \sqrt{5}) = 2 + \sqrt{5} + 4 - 5 + 2 - \sqrt{5} = 3$

$α β γ = 1 (2 + \sqrt{5}) (2 - \sqrt{5}) = - 1$ .

Hence, required cubic equation is $x^{3} - 5 x^{2} + 3 x + 1 = 0$ .

2 + 5 , 1 are roots of the cubic equation given by

Solution