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Find all real numbers x such that $\mid x-14i\mid=1-x$.

The first thing I tried was to just guess an approximate answer and use the calculator to determine that.

If x=-97.5 then,

$$\mid x-14i\mid= \sqrt{(-97.5)^2+14^2}=98.5=1-(-97.5)=1-x$$

But this is only one solution and the questions asks for all real numbers. Maybe this question has only one solution since $14i$ is a constant.

Also, I wanted to show this in a cleaner algebraic fashion instead of just guessing the value.

I assume I have to break up $\mid x-14i\mid$ such that $x$ has an imaginary part equal to zero because the right hand side has no imaginary part but I'm not sure what the method is or what steps to follow to solve this problem algebraically.

Any help would be greatly appreciated.

Thanks

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3 Answers 3

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Well, the thing you need to solve when $x\in\mathbb{R}$ is:

$$\sqrt{x^2+\left(-14\right)^2}=1-x\tag1$$

And that gives only one solution, and that is $x=-\frac{195}{2}$.

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If $x$ is real, then

$$|x-14i|=1-x\implies x^2+196=1-2x+x^2\iff x=-195/2=-97.5$$

It remains to check that $-97.5$ actually is a solution (i.e., that the "$\implies$" goes the other way as well). This is what the OP did.

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\begin{align*} |x-14i|&=1-x\\ |x-14i|^2&=(1-x)^2\\ x^2+14^2&=1+x^2-2x\\ x&=\frac{1-14^2}2 \end{align*} Since it's a "if and only if" chain, the above value for $x$ is the one and only.

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    $\begingroup$ That implies $1-x<0$, so $1-x$ cannot possibly equal $|x-14i|$. $\endgroup$
    – Angina Seng
    Commented May 14, 2020 at 19:22
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    $\begingroup$ You have a sign error in the very last step. It should be $1-14^2$, not $14^2-1$. $\endgroup$
    – Barry Cipra
    Commented May 14, 2020 at 19:24
  • $\begingroup$ thanks both of you $\endgroup$
    – Joe
    Commented May 14, 2020 at 20:35

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