I tried to derive full square but haven't managed to get anything useful from it. The question is: what is (more or less) general method for solving such equations?
Asked
Active
Viewed 97 times
0
-
The quadratic equation still works. – Sep 30 '20 at 22:24
-
@TokenToucan You mean solving it through discriminant? – math-traveler Sep 30 '20 at 22:25
-
3The quadratic formula works over any field where $,2\neq 0$. – Bill Dubuque Sep 30 '20 at 22:33
-
@BillDubuque But what if I get discriminant that is not integer? It seems to be a problem – math-traveler Sep 30 '20 at 22:37
-
The discriminant is $b^2-4ac$: must be an integer. If it’s a nonzero square in $\Bbb Z/17\Bbb Z$, you have two roots. Just do it. – Lubin Sep 30 '20 at 22:40
-
@BillDubuque Well, in this case it isn't and I wonder what to do – math-traveler Sep 30 '20 at 22:44
-
Here the discriminant $\equiv 89\equiv 2^2,$. The usual proof by completing the square shows it has a root in the field $\iff$ the discriminant is a square. This is proved here in many places (including (implcitly) my prior link). – Bill Dubuque Sep 30 '20 at 22:52
-
This provides no insight, but you could also do 17 elementary calculations to solve the equation. – Charles Hudgins Sep 30 '20 at 22:56
1 Answers
0
In this field, the discriminant is a square: $$\Delta=9-4\cdot 5\cdot -4=89\equiv 4\mod 17,$$ hence the roots are $$x_0,x_1=(-3\pm 2)\cdot 10^{-1}=\begin{cases}-5\cdot -5\equiv 8\\ -1\cdot-5=5\end{cases}$$
Completing the square:
First multiply the equation by $5^{-1}=7$. You obtain: $$x^2+4x+6=(x+2)^2+2=(x+2)^2-7^2,$$ whence $x=-2\pm 7$.
J. W. Tanner
- 63,683
- 4
- 43
- 88
Bernard
- 179,256