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