Inspired by Ayush verma
x 8 − x 4 − 2 4 0 = 0
Consider all of the (possibly complex) roots to this equation. How many of them have a positive real part?
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2 solutions
Moderator note:
It was trickly in part because y = x 4 = α > 0 will only yield 1 root with positive real part.
Where does all that knowledge come from?
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Toying around.
Of course, this requires an almost intimate knowledge of roots of unity and their applications.
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I'm slightly surprised at the low correct rate. What do you think tripped people up?
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Is 0 positive? I can see this being confusing for some people.
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@D G – No, 0 is not a positive number, rather it is a non - negative ( or non - positive ) number. The best way to think about it is to consider the set of integers { 1, 2, 3, 4, ... } . This set is the set of all positive Integers and is denoted by m a t h b h Z + . When we add 0 to this, we get the set of Non - Negative Integers ( though there is no standard notation to indicate the set of non - negative integers but still we can use the notation Z + ∪ { 0 } ). In a similar way, we can write the set of negative integers. Adding 0 to this set would yield the set of non - positive integers.
i thought I need to find real root and immediately answered 2. I think this could be a reason for such low rate
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@Aakash Khandelwal – Most of the wrong answers are 2, instead of 4 or 6.
Like me rephrase the question slightly.
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@Calvin Lin – I initially just looked at 1 6 , which had one root with positive real component, then automatically assumed that was true for − 1 5 . But then I decided to look at it again, and found that − 1 5 had two roots with positive real components.
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@Daniel Liu – Can anybody explain me, whether 4th root of a negative number is positive or negative? Please look into given link and help me into this ! http://www.wolframalpha.com/input/?i=x%5E8-x%5E4-240%3D0&dataset=
@Daniel Liu – How do you know if -15 have two positive real component ?
No idea. Maybe it's due to rash decision. Also, it might have been due the other question by Ayush which had the answer de g f / 2 .
How to determine all the complex roots??
( x 4 − 1 6 ) ( x 4 + 1 5 ) = 0 . This yields two groups of four solutions: x 4 = 1 6 ∴ x = ± 2 , ± 2 i , one of which has a positive real parts; and x 4 = − 1 5 ∴ x = 4 1 5 2 ± 1 ± i , two of which have positive real parts.
Therefore the answer is 3 .
I thought 0 is also a real part 😐
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It is-- but it isn't a positive real part, as the problem required.
Another way to deal with the x 4 = − 1 5 is to realize that the 4 roots consist of 2 sets of conjugate pairs, a ± b i and c ± d i . But by Vieta's theorem, the sum of the roots is 0. So a and c have the same numeric value but opposite signs. Therefore one conjugate pair has a positive real part and the other pair has a negative real part, yielding 2 roots in addition to the previous 1 root from x 4 = 1 6 .
Notice that polynomial given is also a polynomial in x 4 . We shall let y = x 4 and we have y 2 − y − 2 4 0 = 0 .
We can easily see that the discriminant is positive; in conjunction with considering the constant term, we have one positive and one negative root exactly. We will call them α and β respectively.
Looking at the case y = x 4 = α > 0 , we have the possible solutions x forming a square with vertices on the real and imaginary axes. In similar fashion, we have the case x 4 = β < 0 with solutions forming a square with edges orthogonal to the axes.
As such, there are 3 solutions strictly to the right of the real axis.