Mark M. answered • 08/20/23
Tutor
5.0
(278)
Mathematics Teacher - NCLB Highly Qualified
it is not hard to show sum (xr,i)2=b2-2c where i is increased from 1 to n
for n=2,,,,xr12+xr22=b2-2c always
Demonstrate
Dayv O.
(x-4)(x+5)=x^2+x-20,,,,b^2-2*c=41=4^2+5^2,,for example,,,demonstrated n=-2
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08/21/23
Dayv O.
f(x)=(x-(1+2i))(x-(1-2i))=x^2-x(1-2i)-x(1+2i)+(1+2i)(1-2i)=x^2-2x+5.,,,b^2-2c=(-)6. (1+2i)^2+(1-2i)^2=1-4+1-4=-6,,,demonstrated n=2
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08/21/23
Dayv O.
can you demonstrate the claim is wrong for n=2 or n=3?
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08/21/23
Roger R.
tutor
The relation is a straightforward consequence of the distributive property. Expand (x−x₁)(x−x₂)⋅ ... ⋅(x−xₙ) and collect the terms (−xₖ)⋅x^(n−1) and (−xₕ)(−xₖ)⋅x^(n−2) where h < k to find b = g(x₁,x₂,...,xₙ) and c = h(x₁,x₂,...,xₙ). Then calculate b²−2c (expand the square).
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08/22/23
Dayv O.
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,agree, in any 1*x^n polynomial as described, b= (-) (xroot1+...+xrootn). Then expand with b^2. Also in any 1*x^n polynomial as described, c= the sum of roots multiplied in pairs. Now xroot1^2+....+xrootn^2=b^2-2c.
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08/22/23
Dayv O.
,,,,,,,for n=2, the simple way to derive formula is just square ((-b/2)+((b^2-4c))^(1/2))/2) and add to squared ((-b/2)-((b^2-4c))^(1/2))/2)=b^2-2c. Higher order proof is really not that hard. Now, do you know of any polynomial contradicting the claim about (b^2)-2c<0 implying at least two roots are complex?08/21/23