题目

求解2元2次方程组!
求解此方程组：
X^2+Y^2=7225
（140-X）^2+（Y-12.5）^2=9506.25

提问时间：2021-03-31

答案

X^2+Y^2=7225
(140-X)^2+(Y-12.5)^2=9506.25
相减
280x-140^2+25y-12.5^2= - 2281.25
280x-25y = 17475
56x-5y = 3495
5y = 3495 - 56x
25x^2+(5y)^2 = 7225*25
25x^2+(3495 - 56x )^2 = 7225*25
3161x^2-391440x [phone]=0
Δ=1062320000
x1 = 391440+32593.25 / 2*3161 =67
x2 = 391440-32593.25 / 2*3161 = 56.76
y的4个值代这里面就行了5y = 3495 - 56x