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I am working on a math problem $(x^{2} - 6x + 45)$ where I must utilize the "completing the square" method, and was able to break down the problem to $(x-3)^{2} = -36.$

I know I have to take the square roots of both sides getting $x-3$ on the left, but for the right, I know 6 is one of the roots for the square root of -36 but what would be the other? Initially, I thought it would be 6i but that is wrong. What is the other root?

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    $\begingroup$ 6 is not a square root of -36. $\endgroup$
    – Nij
    Commented Oct 11, 2021 at 3:19
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    $\begingroup$ 6i is one of them. 6 is not. $\endgroup$
    – Ian
    Commented Oct 11, 2021 at 3:20
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    $\begingroup$ That portal is wrong, and you can verify it directly, $6^2=36 \neq -36$. $\endgroup$
    – Nij
    Commented Oct 11, 2021 at 3:21
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    $\begingroup$ The other one is $-6i$ if you are considering complex solutions. $\endgroup$ Commented Oct 11, 2021 at 3:30
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    $\begingroup$ @SammyBlack Thanks, I was thinking that (or overthinking). I just learned about how I and I^2 translate to sqrt of -1 and -1. So I was confusing myself $\endgroup$
    – יהודה
    Commented Oct 11, 2021 at 3:32

1 Answer 1

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  1. \begin{align}&&x^2&=9\\ &\iff&\sqrt{x^2}&=\sqrt9\\ &\iff&|x|&=3\\ &\iff& x&=\pm3\end{align}

    Notice that the radical sign $\sqrt{\;\;}$ gives just the principal square root; so, $\sqrt {x^2}=|x|\not\equiv x.$ Presumably, this (applying the square root function) is what you mean when you say "to take square root of both sides".

    However, dealing with complex numbers, the definition of "principal root" is arguable: for example, does the principal third root $\sqrt[3] {-1}\;$ of $-1$ equal $$-1\;\text{(real)}$$ or $$\frac12+\frac{\sqrt3}2i=e^{i \frac\pi3}\;\text{(smallest nonnegative argument)}?$$ In the complex world, the convention may still be that the radical sign outputs just the principal root, so it's a good idea to avoid using the symbol $\sqrt[n]{\;\;}$ there.

  2. Just as $$x^2=9\iff x=\pm\sqrt9\:=\pm3,$$ we also have that \begin{align}&&(x-3)^{2} &= -36 \\&\iff &x-3&=\pm6i \\&\iff &x&=3\pm6i.\end{align}

  3. This is incorrect: \begin{align}&&(x-3)^{2} &= -36 \\&\iff& x-3 &= \sqrt{-36}.\end{align}

    This is correct: \begin{align}&&(x-3)^{2} &= -36 \\&\iff& \text{principal square root of }(x-3)^2 &= \sqrt{-36} \\&&&= 6i\\&\iff&x-3&=\pm6i.\end{align}

  4. Also, notice that the various $n$th roots of a number cannot generally be generated from its principal $n$th root by merely adding a $\pm$ in front: \begin{align}&&(x-3)^{3} &= -36 \iff x-3 =-3.30,1.65\pm2.86i.\end{align}

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