If x^2-y= y^2-x=10, then find the value of x*y.
Two parabolas one concave up, one concave to the right.
Should intersect at 4 points.
x^2 - y^2 + x - y = 0
(x-y)(x+y+1)=0
So either x=y
x^2 - x - 10 = 0
x = (1 + √41)/2 or (1 - √41)/2
y = the same
or y = -(x+1)
x^2 + (x+1) - 10 = 0
x^2 + x - 9 = 0
x = (-1 + √37)/2 or (-1 - √37)/2
y = (-1 - √37)/2 or (-1 + √37)/2
(x,y) ........................................ x*y
((1 + √41)/2, (1 + √41)/2) --> (21+√41)/2
((1 - √41)/2, (1 - √41)/2) --> (21-√41)/2
((-1 + √37)/2, (-1 - √37)/2) --> -9
((-1 - √37)/2, (-1 + √37)/2) --> -9
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Posted by Larry
on 2024-09-13 15:01:14 |
Solution