Expert: Scott A Wilson - 11/12/2010

Question
If the sum of the roots is : r + s = -b/a
  The equation is x^2-8x-3=0

What is r^2 + s^2

Answer
The two solutions of a quadratic equation of the from
ax^2 + bx + c = 0 are (-b+sqrt(b^2-4ac))/(2a) and
(-b-sqrt(b^2-4ac))/(2a).

Notice that if both of these roots are added together, only -b/a is left, and that follows with what's given.

Now if both roots are each squared, the numerator will contain
b^2 + 2b*sqrt(b^2-4ac) + b^2-4ac and
b^2 - 2b*sqrt(b^2-4ac) + b^2-4ac,
and the denominator on each of them will be 4a^2..

Adding them together eliminates the middle term and gives
2(b^2 + b^2 - 4ac) = 4(b^2 - ac).

Putting them both together cancels the 4 in each of them,
and this gives (b^2 - 4ac)/a^2.

The final step is to note that a = 1, b = -8, and c = -3,
then put these into the equation in the last paragraph.