Question

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

Answer 1

`alpha^3 + beta^3 = -45/8`

Explanation:

Suppose the roots of the general quadratic equation:

`ax^2+bx+c = 0`

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

`"sum of roots" \ \ \ \ \ \= alpha+beta = -b/a`
`"product of roots" = alpha beta \ \ \ \ = c /a`

So for the given quadratic with roots `alpha` and `beta`:

`2x^2+3x-1 = 0`

we know that:

`alpha+beta = -3/2 \ \ \` ; and `\ \ \ alpha beta = -1/2`

Consider the binomial expression:

`\ \ \ \ \ (alpha+beta)^3 = alpha^3 + 3alpha^2 beta + 3alpha beta^2 + beta^3`
`:. (alpha+beta)^3 = alpha^3 + beta^3+ 3alpha beta(alpha+beta)`
`:. \ alpha^3 + beta^3 = (alpha+beta)^3 - 3alpha beta(alpha+beta)`

Substituting our values of `alpha+beta` and `alpha beta` we get:

`alpha^3 + beta^3 = (-3/2)^3 - 3(-1/2)(-3/2)`
`" " = -27/8 - 9/4`
`" " = -45/8`