Question

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Answer 1

Answer:

x2+2(1+k)x+k2=0

x2+2(1+k)x+k2=(x−a)2

a2+2(1+k)a+k2=x2−2ax+a2

2(1+k)=−2a

k2=a2

k=±a

2(1+a)=−2a

2+2a=−2a

2=−4a

a=−12

2(1−a)=−2a

2−2a=−2a

2≠0

k≠−a

2(1+k)=−2(−12)

2+2k=1

2k=−1

k=−1/2