Need Help Solving 2r^2 +2r+1= 0?

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The equation 2r^2 + 2r + 1 = 0 cannot be factored easily, prompting the use of the quadratic formula to find its roots. The correct application of the formula leads to the calculation of roots as -0.5 ± i. There is a suggestion to ensure proper division and the use of parentheses for clarity in calculations. The discussion emphasizes the importance of understanding complex numbers in the context of quadratic equations. Overall, the conversation centers on solving the quadratic equation accurately.
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I'm feeling really stupid right now, but I can't seem to get this to factor.
2r^2 +2r+1= 0
Can someone please help me out?
 
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You could factor it if you knew its roots, right?
 
so to find the roots I'd use the quadratic formula
-2 +/- \sqrt 4-8 /4
so then would my roots be -.5 +/- 2i?
 
I think you divided wrong. But you have the right idea. (And you should use parentheses when appropriate)
 
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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