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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.
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