2.已知方程 2x – 4(x + 1) + k = 0 有兩個相異實根，求 k 值的範圍。
thc
中4maths
2007-11-19 4:51 am
回答 (2)
2007-11-19 5:40 am
✔ 最佳答案
我估你認該係2x^2 – 4(x + 1) + k = 0唔係2x – 4(x + 1) + k = 0......2x^2 – 4(x + 1) + k = 0 has two unequal real roots
2x^2 – 4(x + 1) + k = 0
2x^2-4x-4+k=0
For Delta: >0
(-4)^2-4(2)(-4+k)>0
16+32-8k>0
48-8k>0
48>8k
6>k
k<6
So,The range of k is <6
參考: Me
2007-11-19 5:32 am
2x – 4(x + 1) + k = 0
2x-4x-4+k=0
-2x+k=0
b^2-4ac>0(real roots)
(-2)^2-4(0)(k)>0
err.. I think the question have problem.(I think so)
2x-4x-4+k=0
-2x+k=0
b^2-4ac>0(real roots)
(-2)^2-4(0)(k)>0
err.. I think the question have problem.(I think so)
收錄日期: 2021-04-23 00:15:03
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