Question 1090501
The quadratic needs to be a perfect square.
{{{(x+a)^2=x^2+2a+a^2}}}
So {{{a^2=1}}}
{{{a^2-1=0}}}
{{{(a+1)(a-1)=0}}}
{{{a=-1}}}
{{{a=1}}}
So then,
{{{k=2a}}}
{{{k=-2}}} and {{{k=2}}}