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Question

$y=x^{10}$ and $y=(x-k)^{10}$ at exactly 1 point where k is a non zero. In terms of k, what is the x coordinate of the intersection

The answer is supposed to be $\frac{k}{2}$ But i only have a slight understanding of how to reach the conclusion

what I am thinking is that we need to get -x=-x+k or something similar because it is an even power but the problem is that is it would give us 0 which is the answer that we dont want to have.

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    Please note that if $x^{2n}=y^{2n}$, then $x=\pm{y}$. Also, if $k=0$, then both function are the same, i.e have infinite points of intersection. – 2017-01-25
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    As it's an even power you either want $x=x-k$ (impossible) or $x=-x+k$ (the solution). You should not get $0$ as an answer at all. – 2017-01-25
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    Notice if $0 \le w < v$ then $w^k < v^k$ and $(\pm x)^{2k} = |x|^{2k}$ so the only way for $w^{2k} = v^{2k}$ is for $|w| = |v|$. And how can $|x| = |x -k|$ if $k \ne 0$? – 2017-01-25
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    " we need to get -x=-x+k or something similar" $-x = -x + k \implies k = 0$ not our hypothesis. The "something similar" is $x = -x + k$ (which is a result of $|x| = |x -k|$) which gives us ... what? – 2017-01-25
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    @JohnRawls consider $a^{10}-b^{10}=(a^5-b^5)(a^5+b^5)$ then $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ and the other similar factorization. – 2017-01-25

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If $|w| \ne |v|$ then $w^{10} \ne v^{10}$. [Because if $0\le|w| < |v|$ then $w^{10}=|w|^{10}< |v|^{10} = v^{10}$ and vice versa.]

So if $x^{10} = (x - k)^{10}$ at one point then there is one $x$ where $|x| = |x -k|$. And $|x| = |x -k|$ only if either $x = x -k$ or $x = -(x-k)$. Now if $k \ne 0$ then $x \ne x-k$ so the only point where $|x| = |x - k|$ is where $x = -(x - k)$ so $ x = -x + k$ so $k = 2x$ so $x = \frac k2$.

$x = \frac k2$ is the only point where $x^{10} = (x - k)^{10}$.