anonymous
  • anonymous
Descartes' Rule of Signs: 2x^4-7x^3+3x^2+8x-4 This equation shows that there would be either 3 positive roots, or 1 positive root, but my answer and the answer at the back of my textbook say that there are only two positive roots; 1/2, and 2. The Descartes' Rule of Signs says that the number of positive roots is either equal to the number of variations in sign or is less than that by an even whole number. Why does that not apply to this equation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
what are the number of variations in your positive setup?
anonymous
  • anonymous
I was thinking that it was 3 variations
amistre64
  • amistre64
2x^4 -7x^3 +3x^2 +8x -4 | | | 0 1 2 3 i count 3 as well

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amistre64
  • amistre64
do you recall that a root can be repeated right?
anonymous
  • anonymous
I was thinking about that also, but why would we have to subtract the number of variations by 2, instead of 1, if we were accounting for repeats?
amistre64
  • amistre64
the subtraction by 2 is to account for complex (nonreal) roots which always come in conjugate pairs
anonymous
  • anonymous
I see. What would we do about the repeating roots, then?
amistre64
  • amistre64
repeating roots was just a thought, but it has 3 postivie roots http://www.wolframalpha.com/input/?i=0%3D2x^4-7x^3%2B3x^2%2B8x-4
amistre64
  • amistre64
does the question ask you to find the roots? or just the number of them?
anonymous
  • anonymous
It asks to find all the real roots.
amistre64
  • amistre64
and you are sure that your looking up the right answer key with the right question? also, books do have errors in them
amistre64
  • amistre64
attaching a picture might help verify your cause :)
anonymous
  • anonymous
Yep, it has the roots 1/2, 2, and -1. Doesn't the graph in the link have two positive roots and one negative root? :o
amistre64
  • amistre64
i am so glad you are on top of these things ... so yeah, there is a double root. so, it has 3 positive roots. 1/2, 2, 2 it is just that one of them occurs more than once.
anonymous
  • anonymous
Okays, so when solving these kinds of problems, we should assume there is a double root if the answer isn't the correct number of variations?
amistre64
  • amistre64
you discover it either by working out the division ... or by the graph
anonymous
  • anonymous
I see. Graphing...
amistre64
  • amistre64
as long as no errors occur in print, yeah, you can assume that one or more of the roots are a multiple root if we count 3 or 1, and only find '2' of them.
anonymous
  • anonymous
Great, thanks!
amistre64
  • amistre64
your welcome

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