more about quadratic equation
Pls help. I just want the formula to do .
or u may help me to solve
Solve the following equation:
1. 6x-√x-2=0
2.If the area and the perimeter of a rectangle are 360cm^2 and 78 com respectively, find the dimensions of the rectangle. (I duno whether there are two equations, but i only want to hv one euation)
3. the sum of a number and the five times of its reciprocal is 6. Find the possible values of the number. (let the x? i don't know wt is "the sum if a number- -""that means two numbers?)
quadratic equation
2008-03-10 6:46 am
回答 (2)
2008-03-10 7:12 am
✔ 最佳答案
1.let y = x^1/26y^2-y-2=0
(3y-2)(2y+1)=0
y=2/3 (y=-1/2 rejected)
x=4/9
2. let x be the length and y be the width.
xy=360
2x+2y=76 x+y=39 x=38-y (i think it should be 76)
(39-y)y=360
39y-y^2=360
y^2-38y+360=0
(y-18)(y-20)=0
y=18 or 20
3. let x be the number,
x+5/x=6
x+5/x-6=0
x^2+5-6x=0
(x-5)(x-1)=0
x=5 or1
2008-03-10 7:13 am
(1)
6x-√x-2=0
(3√x +2) (√x -1) = 0
√x= -2/3 or √x = 1
x = 4/9 or x = 1
(2)
Let x be length of rectangle,
the width of rectangle = (78 - 2x)/2 = 39 - x
x (39 - x) = 360
x^2 - 39x + 360 = 0
(x - 24) (x-15) = 0
x = 24 or x = 15
when x = 24cm, y = 15cm
when x = 15cm, y = 24cm
(3)
Let the number = x
x + 5/x = 6
x^2 + 5x -6 =0
(x + 6) (x-1) =0
x = 1 or x = -6
2008-03-09 23:15:30 補充：
(3) Revised the answer
Let the number = x
x + 5/x = 6
x^2 - 6x + 5 =0
(x - 5) (x-1) =0
x = 1 or x = 5
6x-√x-2=0
(3√x +2) (√x -1) = 0
√x= -2/3 or √x = 1
x = 4/9 or x = 1
(2)
Let x be length of rectangle,
the width of rectangle = (78 - 2x)/2 = 39 - x
x (39 - x) = 360
x^2 - 39x + 360 = 0
(x - 24) (x-15) = 0
x = 24 or x = 15
when x = 24cm, y = 15cm
when x = 15cm, y = 24cm
(3)
Let the number = x
x + 5/x = 6
x^2 + 5x -6 =0
(x + 6) (x-1) =0
x = 1 or x = -6
2008-03-09 23:15:30 補充：
(3) Revised the answer
Let the number = x
x + 5/x = 6
x^2 - 6x + 5 =0
(x - 5) (x-1) =0
x = 1 or x = 5
收錄日期: 2021-04-29 20:27:37
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