The roots of the quadratic 2x^2+3x-1=0 are alpha and beta. Without calculating the roots, find alpha^3+beta^3?

Apr 1, 2017

${\alpha}^{3} + {\beta}^{3} = - \frac{45}{8}$

Explanation:

Suppose the roots of the general quadratic equation:

$a {x}^{2} + b x + c = 0$

are $\alpha$ and $\beta$ , then using the root properties we have:

$\text{sum of roots} \setminus \setminus \setminus \setminus \setminus \setminus = \alpha + \beta = - \frac{b}{a}$
$\text{product of roots} = \alpha \beta \setminus \setminus \setminus \setminus = \frac{c}{a}$

So for the given quadratic with roots $\alpha$ and $\beta$:

$2 {x}^{2} + 3 x - 1 = 0$

we know that:

$\alpha + \beta = - \frac{3}{2} \setminus \setminus \setminus$ ; and $\setminus \setminus \setminus \alpha \beta = - \frac{1}{2}$

Consider the binomial expression:

$\setminus \setminus \setminus \setminus \setminus {\left(\alpha + \beta\right)}^{3} = {\alpha}^{3} + 3 {\alpha}^{2} \beta + 3 \alpha {\beta}^{2} + {\beta}^{3}$
$\therefore {\left(\alpha + \beta\right)}^{3} = {\alpha}^{3} + {\beta}^{3} + 3 \alpha \beta \left(\alpha + \beta\right)$
$\therefore \setminus {\alpha}^{3} + {\beta}^{3} = {\left(\alpha + \beta\right)}^{3} - 3 \alpha \beta \left(\alpha + \beta\right)$

Substituting our values of $\alpha + \beta$ and $\alpha \beta$ we get:

${\alpha}^{3} + {\beta}^{3} = {\left(- \frac{3}{2}\right)}^{3} - 3 \left(- \frac{1}{2}\right) \left(- \frac{3}{2}\right)$
$\text{ } = - \frac{27}{8} - \frac{9}{4}$
$\text{ } = - \frac{45}{8}$