Question

若方程8x +2kx+k-1=0的两个实数根是x1 ,x 2且满足x 1^2+x2^2 =1,则k的值为( ).A.-2

Answer 1

由韦达定理,得

x1+x2=-k/4

x1x2=(k-1)/8

x1^2+x2^2

=(x1+x2)^2-2x1x2

=k^2/16-(k-1)/4

=(k^2-4k+4)/16

=(k-2)^2/16=1

(k-2)/4=1或(k-2)/4=-1

k=6或k=-2

k=6代回去验证：8x^2+12x+5=0,判别式