1)find thw values of x: x^2-9x+32>=0
----------
2)solve: < 3x+y=0
3x^2+4y-3=0
簡單數學題目一問
2010-05-16 6:46 pm
回答 (2)
2010-05-16 7:48 pm
2010-05-20 7:20 pm
1) a= 1, b=-9 , c =32
△=b^2-4ac
=(-9)^2-4(1)(32)
=-47
<0,
so, there are no intersection point with x-axis,
as a>1, the curve should be a u shape opening upwards,
so the solution of x^2-9x+32>=0 should be all real no.
2)
3x+y=0
y=-3x -------(1)
3x^2+4y-3=0 ------ (2)
sub (1) into (2)
3x^2 +4(-3x) -3 = 0
3x^2 -12x-3=0
x= 4.24 or -0.24
△=b^2-4ac
=(-9)^2-4(1)(32)
=-47
<0,
so, there are no intersection point with x-axis,
as a>1, the curve should be a u shape opening upwards,
so the solution of x^2-9x+32>=0 should be all real no.
2)
3x+y=0
y=-3x -------(1)
3x^2+4y-3=0 ------ (2)
sub (1) into (2)
3x^2 +4(-3x) -3 = 0
3x^2 -12x-3=0
x= 4.24 or -0.24
收錄日期: 2021-04-25 13:26:09
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