# Are Complex Solutions Overlooked When Solving $$16x^4 = 81$$ Directly?

• RChristenk
RChristenk
Homework Statement
Solve ##16x^4=81##
Relevant Equations
Algebra
##x^4=\dfrac{81}{16}##

##x=\pm\dfrac{3}{2}##.

But I recently realized there are complex solutions as well:

##16x^4-81=0##

##(4x^2)^2-9^2=0##

##(4x^2+9)(4x^2-9)=0##

##x^2=\dfrac{-9}{4}, x^2=\dfrac{9}{4}##

##x=\pm\dfrac{3i}{2}, x=\pm\dfrac{3}{2}##

Intuitively when I see ##16x^4=81##, I see a straightforward solution of ##x=\pm\dfrac{3}{2}##. But solving problems in this way clearly excludes the complex solutions. Why is that so? Does it mean this straightforward approach to solving these kinds of problems is wrong?

$$x^4=(\frac{3}{2})^4$$
$$x^2=\pm(\frac{3}{2})^2$$
$$x=\pm\frac{3}{2},\pm \frac{3}{2}i$$
For complex number x, I do not find any concern here. Quartic equation has four roots.

$$x^4=1$$
$$x=e^{\frac{n\pi}{2}i}\ \ ,n=0,1,2.3$$

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MatinSAR, DaveE and SammyS
RChristenk said:
Homework Statement: Solve ##16x^4=81##
Relevant Equations: Algebra

##x^4=\dfrac{81}{16}##

##x=\pm\dfrac{3}{2}##.

But I recently realized there are complex solutions as well:

##16x^4-81=0##

##(4x^2)^2-9^2=0##

##(4x^2+9)(4x^2-9)=0##

##x^2=\dfrac{-9}{4}, x^2=\dfrac{9}{4}##

##x=\pm\dfrac{3i}{2}, x=\pm\dfrac{3}{2}##

Intuitively when I see ##16x^4=81##, I see a straightforward solution of ##x=\pm\dfrac{3}{2}##. But solving problems in this way clearly excludes the complex solutions. Why is that so? Does it mean this straightforward approach to solving these kinds of problems is wrong?
Did not the problem define what kind of object is ##x##, i.e., real number, complex number, something else?

RChristenk said:
Homework Statement: Solve ##16x^4=81##
Relevant Equations: Algebra

##x^4=\dfrac{81}{16}##

##x=\pm\dfrac{3}{2}##.

But I recently realized there are complex solutions as well:

##16x^4-81=0##

##(4x^2)^2-9^2=0##

##(4x^2+9)(4x^2-9)=0##
If you see a pattern ##x^{even} - a^2 = 0## then you can immediately start with what you found anyway here. And you should go on!

\begin{align*}
0&=16x^4-81=(4x^2+9)(4x^2-9)=(4x^2+9)(2x+3)(2x-3)\\&=(2x+3 i )(2x - 3 i) (2x+3)(2x-3)
\end{align*}
This version of a binomial formula is very often helpful. It should become a reflex. It is not hard to use and doesn't waste much time if it won't help.

dextercioby and mathwonk
RChristenk said:
But I recently realized there are complex solutions as well:
Good catch!
RChristenk said:
##16x^4-81=0##

##(4x^2)^2-9^2=0##

##(4x^2+9)(4x^2-9)=0##

##x^2=\dfrac{-9}{4}, x^2=\dfrac{9}{4}##

##x=\pm\dfrac{3i}{2}, x=\pm\dfrac{3}{2}##

Intuitively when I see ##16x^4=81##, I see a straightforward solution of ##x=\pm\dfrac{3}{2}##.
Yes, and that is what would be expected if complex numbers have not been taught yet or if, as @Hill said, x has been specified as a real number.
RChristenk said:
But solving problems in this way clearly excludes the complex solutions. Why is that so? Does it mean this straightforward approach to solving these kinds of problems is wrong?
What you call the "straightforward approach" is just the result of not having complex solutions as a familiar part of your intuition. I don't know what was expected in your class, but you are far ahead if you instinctively realize that there are complex solution. Once you recognize that fact, you can decide if a particular problem should accept the complex solutions. Often the complex solutions do not make sense as a real-world answer, but other times they do (especially if oscillations, cyclic behavior, or frequencies are involved).

RChristenk
Well, if it's any consolation, the FTA guarantees you you've found all roots now.

PhDeezNutz and FactChecker

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