x + y = 20 x*y = 44 y = 20 - x x*(20-x) = 44 20*x - x^2 = 44 x^2 - 20*x + 44 = 0 Note that this quadratic contains essentially the given problem - the roots of the quadratic are the numbers we seek. Because of symmetry, it doesn't matter which is which. The result is x = 10 + 2*sqrt(14), y = 10 - 2*sqrt(14) or vice versa.
Why does no one use this simple substitution method? Substitute x=10+z and y=10-z into the second equation and rearrange to 100-z^2=44 or z^2=56, which has roots z=±2√14. Thus x=10±2√14 and y=10∓2√14.
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x + y = 20
x*y = 44
y = 20 - x
x*(20-x) = 44
20*x - x^2 = 44
x^2 - 20*x + 44 = 0
Note that this quadratic contains essentially the given problem - the roots of the quadratic are the numbers we seek. Because of symmetry, it doesn't matter which is which. The result is x = 10 + 2*sqrt(14), y = 10 - 2*sqrt(14) or vice versa.
Why does no one use this simple substitution method?
Substitute x=10+z and y=10-z into the second equation and rearrange to 100-z^2=44 or z^2=56, which has roots z=±2√14.
Thus x=10±2√14 and y=10∓2√14.