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Comments
I know it's a little late but I'll post my solution, the equation is equivalent to
x3−5x2+8x−4+6(x−3x2−x+1)=0⇔(x−1)(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x3−x2+x−1)=0⇔(x−1)(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x−1)(x2+1)=0⇔(x−1)[(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x2+1)]=0
We see that (x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x2+1)>0;∀x∈R so the only root is x=1
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
I know it's a little late but I'll post my solution, the equation is equivalent to x3−5x2+8x−4+6(x−3x2−x+1)=0 ⇔(x−1)(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x3−x2+x−1)=0 ⇔(x−1)(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x−1)(x2+1)=0 ⇔(x−1)[(x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x2+1)]=0 We see that (x−2)2+x2+x3x2−x+1+3(x2−x+1)26(x2+1)>0;∀x∈R so the only root is x=1
Thanks so much!
May be I can help. But what is the exponent of x2−x+1 ?
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1/3