设抛物线y2=4x与直线y=2x+k相交所得弦长为|AB|=3根号5.（1）求k

将直线y=2x+k带入y^2=4x,
∴4x^2+(4k-4)x+k^2=0
设两点的横坐标是x1,x2
相应的纵坐标为2x1+k,2x2+k
∵│AB│=3√5,
∴3√5=√[(x1-x2)^2+(y1-y2)^2]
=√((x1-x2)^2+(2x1-2x2)^2)
=√5(x1-x2)^2
=√(5(x1+x2)^2-20x1x2)
∵x1+x2=k-1,
又∵x1x2=k^2/4
∴3√5=√(5*(k-1)^2-20*k^2/4)
∴k=-4