Revision history [back]
 | 1 | initial version |
I think you are looking for the roots of the polynomial:
f = x^2 - 30*x + 2817
f.roots()which gives:
[(-36*I*sqrt(2) + 15, 1), (36*I*sqrt(2) + 15, 1)]This would mean that your original function is equal to:
x2−30∗x+2817=(x−(15−36√(−2))(x−(15+36√(−2))
| | 2 | No.2 Revision |
I think you are looking for the roots of the polynomial:
f = x^2 - 30*x + 2817
f.roots()which gives:
[(-36*I*sqrt(2) + 15, 1), (36*I*sqrt(2) + 15, 1)]This would mean that your original function is equal to:
$$ x^2-30*x+2817 x^2-30x+2817 = \left(x-(15-36 \sqrt(-2)\right)\left(x-(15+36\sqrt(-2)\right)$$
| | 3 | No.3 Revision |
I think you are looking for the roots of the polynomial:
f = x^2 - 30*x + 2817
f.roots()which gives:
[(-36*I*sqrt(2) + 15, 1), (36*I*sqrt(2) + 15, 1)]This would mean that your original function is equal to:
$$ x^2-30x+2817 = \left(x-(15-36 \sqrt(-2)\right)\left(x-(15+36\sqrt(-2)\right)$$\sqrt{-2}\right)\left(x-(15+36\sqrt{-2}\right)$$
| | 4 | No.4 Revision |
I think you are looking for the roots of the polynomial:
f = x^2 - 30*x + 2817
f.roots()which gives:
[(-36*I*sqrt(2) + 15, 1), (36*I*sqrt(2) + 15, 1)]This would mean that your original function is equal to:
$$ x^2-30x+2817 = \left(x-(15-36 \sqrt{-2}\right)\left(x-(15+36\sqrt{-2}\right)$$\sqrt{-2})\right)\left(x-(15+36\sqrt{-2})\right)$$
