Find two distinct numbers, so that
x^2-y=91
y^2-x=91
Two parabolas could have 0,1,2,3, or 4 intersections.
x^2 - y^2 + x - y = 0
(x-y)(x+y+1) = 0
case 1: x=y
x^2 - x - 91 = 0
x = (1 ± √365)/2
y = same as x
case 2: x = -(y+1) --> y = -(x+1)
x^2 + (x+1) - 91 = 0
x^2 + x - 90 = 0
x = (-1 ± √361)/2
x = (-1 ± 19)/2
x = {-10, 9} with y = -(x+1)
y = {9, -10}
So the four solutions are:
((1 + √365)/2,(1 + √365)/2)
((1 - √365)/2,(1 - √365)/2)
(-10, 9)
(9, -10)
------
little program to check the results:
a = (1 + (365)**.5)/2
b = (1 - (365)**.5)/2
solutions = [ [a,a], [b,b], [-10, 9], [9,-10] ]
def f(x,y):
return x**2 - y , y**2 - x
for d in solutions:
print(f(d[0],d[1]))
Output:
(90.99999999999999, 90.99999999999999)
(91.0, 91.0)
(91, 91)
(91, 91)
|
|
Posted by Larry
on 2023-04-06 11:21:15 |
Solution
