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

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The roots of the equation x^2+ax=2 are x1 and x2.

=> x1^2 + ax1 = 2 and x2^2 + ax2 = 2

Adding the two x1^2 + x2^2 + a(x1 + x2) = 4

But it is given that x1^2 + x2^2 = 4, this gives a = 0

**The required value of a is 0**

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.**