if alpha (α) and beta (β) are the roots of 2x^2-7x+10=0, find: (α) + (β) (α) x (β) (α)^2 + (β)^2 (α)^2 x (β) + (β)^2 x (α)

We'll find the roots of the quadratic using the
formula:

`x_1 = (-b+sqrt(b^2 -
4ac))/(2a)`

`x_2 = (-b - sqrt(b^2 -
4ac))/(2a)`

a,b,c are the coefficients of the
quadratic:

a = 2 ; b = -7 and c =
10

`x_1 = (7 + sqrt(49 -
80))/(4)`

We notice that the result of the difference under
the radical sign is negative, therefore the equation has complex
roots.

`x_1 = (7+isqrt31)/4` and `x_2 = (7 -
isqrt31)/4`

We'll put `x1 = alpha` and `x_2 =
beta`

`alpha + beta = (7 + isqrt31 + 7 -
isqrt31)/4`

We'll eliminate imaginary
parts:

`alpha + beta =
14/4`

`alpha + beta =
7/2`

We'll calculate the product of roots, that is a
special product which returns a difference of two
squares:

`alpha*beta = (49 -
31i^2)/16`

But `i^2 = -1` => `alpha*beta =
(49+31)/16` => `alpha*beta = 5`

We'll calculate
`alpha^2 + beta^2` :

`alpha^2 + beta^2 = (alpha + beta)^2 -
2alpha*beta`

`alpha^2 + beta^2 = 49/4 - 10` =>
`alpha^2 + beta^2 = 9/4`

We'll calculate `alpha^2*beta +
beta^2*alpha = alpha*beta(alpha + beta)`

`alpha^2*beta +
beta^2*alpha = 5*7/2 = 35/2`

Therefore, the
requested results are: `alpha + beta = 7/2 ; alpha*beta = 5 ; alpha^2 + beta^2 = 9/4 and
alpha^2*beta + beta^2*alpha = 35/2` .