Question #aa189

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Answer 1 · 2017-04-12T13:35:36

$k=0.$

$p(x)=x^2-x-2.$

To find $k$ , such that, $AA x,p(x+k)=p(x-k).$

Now, $p(x+k)=p(x-k),$

$rArr (x+k)^2-(x+k)-2=(x-k)^2-(x-k)-2,$

$rArr x^2+2kx+k^2-x-k-2=x^2-2kx+k^2-x+k-2,$

$rArr 4kx-2k=0, or, 2k(2x-1)=0$

$rArr k=0.$