已知y1=x2﹣2x+3,y2=3x﹣k.
(1)当x=1时,求出使等式y1=y2成立的实数k;
(2)若关于x的方程y1+k=y2有实数根,求k的取值范围.
解:(1)当x=1时,y1=2,y2=3﹣k,
根据题意,得:2=3﹣k,
解得:k=1;
(2)由题意,x2﹣2x+3+k=3x﹣k,即x2﹣5x+3+2k=0有实数根,
∴△=(﹣5)2﹣4(3+2k)≥0,
解得:k≤
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A: ____
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关于电场线,以下说法正确的是( )