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I’m doing a study prep, and was faced with the problem enclosed below. I’m not sure how to go about, the answer is D. If someone could post a step by step that would be massively helpful.

Solve the equation in the complex number system: $x^4 - 6x^2 - 7= 0$

(a) $\{ \sqrt{7} i, i \}$

(b)$\{ -\sqrt{7} i, -i \}$

(c) $\{ \sqrt{7}, 7 \}$

(d) $\{ -\sqrt{7}, \sqrt{7}, i, -i \}$

Edit: This problem is now solved by using $y=x^2.$

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    $\begingroup$ Welcome to Mathematics Stack Exchange. First let $y=x^2$ and solve for $y$ $\endgroup$
    – J. W. Tanner
    Commented Sep 6, 2023 at 2:40
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    $\begingroup$ Since you don't know how to solve a general 4th degree equation, perhaps this is a special example where there is a better way... $\endgroup$
    – GEdgar
    Commented Sep 6, 2023 at 2:42
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    $\begingroup$ Tanner, thank you so much, that was the step I couldn’t remember for the life of me! $\endgroup$
    – Ezra Nielsen
    Commented Sep 6, 2023 at 2:52
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    $\begingroup$ I don’t need to solve it, but it’s a test prep so I I need to know how to do it, either way thanks to Tanner now I do $\endgroup$
    – Ezra Nielsen
    Commented Sep 6, 2023 at 2:53
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    $\begingroup$ Does this answer your question? How do I find the roots of a quartic, without guessing? $\endgroup$
    – Anne Bauval
    Commented Sep 6, 2023 at 3:21

2 Answers 2

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Without solving it directly via a quadratic substitution you can determine the polynomial must have 4 complex roots with respect to multiplicity by the Fundamental Theorem of Algebra. Since all the coefficients of the quartic are real, we also know if any complex number is a solution, its conjugate must also be a solution. This only leaves D.

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Sometimes, the simplest solution is the "best" solution.

Notice that the powers of $x$ are all even. But then you get the same value if you plug in $x= a$ into $x^4 - 6x^2 - 7$ as you would if you plugged in $x= -a$. But then if you have a solution, say $S$, then $-S$ is also a solution.

So the solutions come in plus or minus pairs (unless one of them happens to be zero, which from the answers, cannot be the case here). That immediately gives the answer out of the four choices. Can you see why and which?

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  • $\begingroup$ Yeah, the problem is that this is just a test prep so it’s the methodology that meters the most, the actual test isn’t multiple choice. $\endgroup$
    – Ezra Nielsen
    Commented Sep 6, 2023 at 2:56

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