x^2+(x+1)^2=1252?
回答 (4)
2008-07-23 3:27 pm
✔ 最佳答案
24.515
參考: (x+1)^2 = (x+1)*(x+1) = (x^2 + 2x + 1) so,
x^2 + (x^2 + 2x + 1) = 1252 or
2x^2 + 2x = 1252 - 1
now remove the subtraction and divide by 2, you get:
(2x^2 + 2x)/2 = 1251/2 -->
x^2 + x = 625.5
2008-07-23 10:17 pm
x^2+ (x^2+2x+1) = 1252
2x^2+2x+1=1252
2(x^2+x)=1251
x^2+x=625.5
x^2+x-625.5=0
This is the quadratic equation ax^2+bx+c=0
Using the calculator, you can get that:
x= 28.18928344 or -22.18928344
2x^2+2x+1=1252
2(x^2+x)=1251
x^2+x=625.5
x^2+x-625.5=0
This is the quadratic equation ax^2+bx+c=0
Using the calculator, you can get that:
x= 28.18928344 or -22.18928344
2008-07-23 10:11 pm
What????????
2008-07-23 10:09 pm
x^2+x^2 + 2x + 1 = 1251
2x^2 + 2x +1-1251 =0
2x^2 + 2x -1250 =0
2(x^2 +1x -625)=0
go from there
2x^2 + 2x +1-1251 =0
2x^2 + 2x -1250 =0
2(x^2 +1x -625)=0
go from there
收錄日期: 2021-04-19 01:31:41
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