Question

If the sum of the roots of the quadratic equation is 3 and sum of their cubes is 63, find the quadratic equation.

Answer 1

Let α and β be the roots of the quadratic equation. Then, α + β = 3 and α³ + β³ = 63.
On using the identity (a³ + b³) = (a + b)³ – 3ab(a + b), we get:
(α³ + β³) = (α + β)³ – 3αβ(α + β)
On substituting α + β = 3 and α³ + β³ = 63, we get:
63 = (3)³ – 3αβ(3)
$\Rightarrow$ 63 = 27 – 9αβ
$\Rightarrow$ 9αβ = 27 – 63
$\Rightarrow$ 9αβ = –36
$\Rightarrow$ αβ = $\frac{-36}{9}=-4$
We know that if α and β are the roots of a quadratic equation, then the quadratic equation is
x² – (α + β)x + αβ = 0
On substituting α + β = 3 and αβ = –4, we get:
x² – (3)x + (–4) = 0
$\Rightarrow$ x² – 3x – 4 = 0