Given the equation x^2+ax=2 and x1, x2 the solutions of the equation what is a if x1^2+x2^2=4 ?

We'll use the Viete's relations to express the sum and the
product of the roots.

x1 + x2 =
-b/a

x1*x2 = c/a

a,b,c, are
the coefficients of the equation.

We'll re-write the given
equation, moving all terms to one side:

x^2 + ax - 2 =
0

a = 1 , b = a and c = -2

x1
+ x2 = -a/1 (1)

x1*x2 = -2/1
(2)

x1^2 + x2^2 = (x1 + x2)^2 -
2x1*x2

But, from enunciation, x1^2 + x2^2 =
4

4 = (x1 + x2)^2 - 2x1*x2
(3)

We'll substitute (1) and (2) in
(3):

4 = (-a)^2 - 2*(-2)

4 = a
+ 4

We'll subtract 4 both
sides:

a =
0.

So, the value of "a", for the sum of the
squares of the roots of the equation to be 4 is: a =
0.