# Calculate x1^2 + x2^2. x1 and x2 are the solution of the equation x^2-2x+5=0

*print*Print*list*Cite

x^2 -2x+5=0

Since x1 and x2 are solutions to the equation:

==> x1^2 -2x1 +5 =0.....(1)

and x2^2 -2x^2+5 =0 ....(2)

We will add (1) and (2) :

x1^2 -2x1+5 + x2^2 -2x2 +5 =0

Rearrange:

x1^2+ x2^2 = 2(x1+x2) -10

But according Viete's rule x1+x2 = -b/a = 2

==> x1^2 +x2^2 = 2(2)-10 = -6

x^2-2x+5 = 0. To find x1^2+x2^2. where x1 and x2 are the roots of the given equation.

Solution:

By the the relation between the roots and coefficients,

x1+x2 = -coefficient of x/coeeficient of x^2 = -(-2)/1 =2.

x1x2 = constant term/coefficient of x^2 = 5/1 =5.

Therefore,

x1^2+x2^2 = (x1+x2)^2 -2x1x2= 2^2 - 2(5) =4-10 = -6.

We'll substitute the roots into the equation, so that:

x1^2-2x1+5=0 (1)

x2^2-2x2+5=0 (2)

We'll add the relations (1)+(2):

x1^2-2x1+5+x2^2-2x2+5=0

x1^2+x2^2 = 2(x1+x2) - 10

But, x1+x2 = 2 (From Viete's relations)

x1^2+x2^2 = 2*2 - 10

x1^2+x2^2 = 4-10

**x1^2+x2^2 = -6**