let a and b be the solution of 3x^2-x-3=0.Then find a^2+b^2

since 3x^2 - x - 3 is not factorable, we will have to
solve the equation 3x^2 - x - 3 = 0 using the quadratic
formula.

the quadratic formula
is:

let ax^2 + bx + c = 0

x =
[-b + sqrt(b^2 - 4ac)]/2a   or  x = [-b - sqrt(b^2 -
4ac)]/2a

with the equation 3x^2 - x - 3 = 0, we
get:

                                a = 3, b = -1, c =
-3

using this, substitution gives us the following
expressions for x:

x(1) = [-(-1)+sqrt((-1)^2 -
4(3)(-3))]/2(3)

       = [1+sqrt(1 +
36)]/6

       = [1 +
sqrt(37)]/6

x(2) = [-(-1)-sqrt((-1)^2 -
4(3)(-3))]/2(3)

       = [1-sqrt(1 +
36)]/6

       = [1 -
sqrt(37)]/6

since x(1) and x(2) are the solutions to the
equation, let a = x(1) and b = x(2).

a =
[1+sqrt(37)]/6

a^2 =
{[1+sqrt(37)]/6}^2

       =
[1+sqrt(37)]^2/6^2

       =
[1+2sqrt(37)+sqrt(37)^2]/36

       =
[1+2sqrt(37)+37]/36

       =
[38+2sqrt(37)]/36

       =
[19+sqrt(37)]/18

b =
[1-sqrt(37)]/6

b^2 =
{[1-sqrt(37)]/6}^2

      =
[1-sqrt(37)]^2/6^2

      =
[1-2sqrt(37)+sqrt(37)^2]/36

      =
[1-2sqrt(37)+37]/36

      =
[38-2sqrt(37)]/36

      =
[19-sqrt(37)]/18

a^2 + b^2 = [19+sqrt(37)]/18 +
[19-sqrt(37)]/18

                 =
[19+sqrt(37)+19-sqrt(37)]/18

                 = [19 +
19]/18

                 =
38/18

                 = 19/9   or 2
1/9