How many complex roots does the equation below have?

Mathematics

1 answer:

Maslowich3 years ago

4 0

Answer:

The number of complex roots is 6.

Step-by-step explanation:

Descartes's rule of signs tells you that the number of positive real roots is 0. The number of negative real roots will be at most 2. The minimum value of the left side will be between x=0 and x=-1, but will never be negative. Thus all six roots are complex.

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The magnitude of x^3 will exceed the magnitude of x^6 only for values of x between -1 and 1. Since the magnitude of either of these terms will not be more than 1 in that range, the left-side expression must be positive everywhere.

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MissTica

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I actually began with the guess-and-check method, but seeing as that won't always work, let's go over the formal way. To get the original number, you first need to determine how many times the number was squared.

To make it simple, let's use x to focus on the exponents. The number was squared 3 times, so x^2, x^2, x^2. Basically, you need to multiply. 2 * 2 * 2 = 8. So, now find the 8th root of 6561 (depending on the calculator, you can just input it). You should come up with 3. Let me know if this part confuses you.

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4 years ago

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crimeas [40]

It would be 16 more 6th graders that are not in either band or orchestra.
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