Solve Quadratic Equation for x1^2 + x2^2

AI Thread Summary
To find x1^2 + x2^2 for the equation x^2 - 2x + 4 = 0, Vieta's theorem indicates that x1 + x2 = 2 and x1 * x2 = 4. Since the roots are complex, solving for x directly is not feasible. The relationship x1^2 + x2^2 can be derived using the equation (x1 + x2)^2 - 2x1x2. By expanding (x1 + x2)^2, the calculation can be completed to find the desired sum. The discussion emphasizes the importance of using Vieta's theorem in cases with complex roots.
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Homework Statement



If the roots of x^2 - 2x + 4 = 0 are x1 and x2, what is x1^2 + x2^2?

Homework Equations





The Attempt at a Solution



I don't think these questions can be answered by solving for x only cause there will be no real root(complex number).

So I found that x1 + x2 = 2 and x1 * x2 = 4 via Vieta's theorem.
And I tried solving it with the quadratic formula but as I said it hasn't got a real root.
 
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blunted said:

Homework Statement



If the roots of x^2 - 2x + 4 = 0 are x1 and x2, what is x1^2 + x2^2?

Homework Equations





The Attempt at a Solution



I don't think these questions can be answered by solving for x only cause there will be no real root(complex number).

So I found that x1 + x2 = 2 and x1 * x2 = 4 via Vieta's theorem.
And I tried solving it with the quadratic formula but as I said it hasn't got a real root.

That's a good start.

x12 + x22 = (x1 + x2)2 - 2x1x2, right?
 
How did you get to that equation? :O
 
blunted said:
How did you get to that equation? :O
Expand (x1 + x2)2 & see what you get.
 
Oh, right.. Thanks!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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