Given:
x+ sqrt(y)=7
sqrt(x)+y =11
Solve for integer values of x & y, formally, neither by guessing nor software.
Assume that x is less than y.
2nd method: assume x&y are integers and x<y
therefore y = x+a
x + sqrt(y)=7
sqrt(x) + y =11
x + sqrt(x+a)=7
sqrt(x) + x+a =11
sqrt(x+a)=7-x --> x+a = 49 - 14x + x^2
x^2 - 15x + 49-a = 0
x = [15 +/- sqrt(225 + 4a - 196)]/2
x = [15 +/- sqrt(29 + 4a)]/2 a can be {5, 13, 23, ...}
try a=5:
x = (15 + 7)/2 = 11 or x = (15 - 7)/2 = 4
x = 4, y = x+a = 9 satisfies both equations
x = 11, y = x+a = 16 does not
try a=13:
x = (15 + 9)/2 = 12 or x = (15 - 9)/2 = 3
x = 12, y = x+a = 25 does not
x = 3, y = x+a = 16 does not
Larger values of a will produce larger values of either x>7 or y>11
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Posted by Larry
on 2022-03-26 10:03:12 |
2nd method
