Question

Problem for fun:
Solve the system of equations:
a + b + c = 4
ab + bc + ac = -17
abc = -60

Answer 1

substitution will work here.

Answer 2

looks like the roots of a cubic :P

Answer 3

P(x) = mx^3 +nx^2 +wx + z

Answer 4

a = 5, b = 3, and c = -4

Answer 5

actually, a, b, and c, all three could equal 5, 3, and -4.

Answer 6

From the third equation,\[a \neq 0.\]Then,
\begin{eqnarray*}a(b+c) + bc&=&-17 \\
a(4-a)+bc&=&-17 \\
a(4-a)-\frac{60}{a}&=&-17\\
a^3-4a^2-17a+60&=&0.\end{eqnarray*}
Because this monic polynomial with integer coefficients, any rational solution (if it exists) must be an integer. This integer must also divide 60, so if a is rational, \[a \in \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 10, \pm 15, \pm 20, \pm 30, \pm 60\}.\]Here I didn't find an elegant way, so I tested a few of these numbers and found a = -4 is a solution. Then, using long division (or Ruffini's algorithm or something similar) we can find the other two solutions, a = 3 and a = 5.

Answer 7

very nice

Answer 8

Nice indeed, thats what i was trying to get at. There was no difference in solving that system of solving the polynomial:
\[x^{3} -4x^{2} -17x+60 = 0\]
Thats because if i have a cubic polynomial with roots r, s, and t, then:
\[(x - r)(x-s)(x-t) = x^{3}-(r+s+t)x^{2}+(rs+st+rt)x-(rst)\]