# System of two quadratic equations

I'm having hard time figuring out a method for dealing with this type of system of equations. I know they have nice solution, the following system has a solution $p=2$ and $q=1$, $p=\frac{22}{5}$ and $q=\frac{-31}{5}$

$$\left\{ \begin{array}{c} -4p^2-q^2+4pq-18p-36q+81=0 \\ -p^2-4q^2-4pq+8p-4q+4=0 \end{array} \right.$$

I know one way is to factorise them both into two trinomials and get something like : $(ax+by+c)(dx+ey+f)=0$ which then can be combined in system of linear equations (up to 4). I also tried solving a quadratic equation in terms of p or q but ends nothing close to solvable.

• This is a pair of parabolas, so trying to factor into a product of linear terms isn’t going to get you very far. – amd May 24 '18 at 21:09
• Yeah I'm aware of that. Which is why I mentioned it as a potential method to solve which doesn't work every time. I'm looking for something general. – DreaDk May 24 '18 at 21:14
• I don't know if this will help. But adding both will give a circle: $(p+1)^2 + (q+4)^2 = (\sqrt{34})^2$ – sku May 24 '18 at 21:16
• In the general case, you have to solve a quartic equation, which with some work can be reduced to a cubic. Two conics can have up to four intersections. – amd May 24 '18 at 21:16
• In this particular case, you can find a linear combination of the two equations that does factor into linear terms, and then compute their intersections with the circle noted by @sku. – amd May 24 '18 at 21:16

$$(p+1)^2 + (q+4)^2 = 34$$

$\longrightarrow$ a circle.

Subtract both equations:

$$\left(\frac{p+11}{3}-q\right)\left(7-3p-q\right)=0$$

$\longrightarrow$ two lines.

Now the intersection of these two sets is trivial to compute. The line $q = (p+11)/3$ does not intersect the circle. However, at the two points you mentioned, the line $q = 7-3p$ intersects the circle. All this can be determined algebraically.

$\left\{ \begin{array}{c} -4p^2-q^2+4pq-18p-36q+81=0 \\ -p^2-4q^2-4pq+8p-4q+4=0 \end{array} \right.$

Just to noodle about. $-4p^2-q^2+4pq-18p-36q+81 = -(2p -q)^2 -18(p+ 2q) + 81=0$

And $-(p+2q)^2 +4(2p-q) + 4= 0$

So $(2p-q)^2 + 4(2p-q) + 4 = (p+2q)^2 - 18(p+2q) + 81 = (2p-q)^2 + (p+2q)^2$

$(2p-q +2)^2 = (p+2q - 9)^2 = 5(p^2 + q^2)$

$(2p - q + 2) = \pm (p+2q - 9)$

Case 1:

$2p - q + 2 = p+2q - 9$

$p = 3q - 11$

So

$(5q - 20)^2=5((3q-11)^2 +q^2)$

$25q^2 - 200q + 400)= 5(10q^2-66q + 121)$

$25q^2 -130q +205=0$

But $130^2 - 4*205*25 < 0$

Case 2:

$2p - q + 2 = -p-2q + 9$

$q = -3p +7$

So

$(5p -5)^2=5(p^2 + (-3p+7)^2)$

$25p^2 - 50p + 25 = 5(10p^2 -42p + 49)$

$25p^2 -160p +220 =0$

$5p^2 - 32p + 44 = 0$

$p =\frac {32 \pm\sqrt{32^2 - 44*4*5}}{10}$

$= \frac {32 \pm 4{\sqrt 9}}{10} =\frac {32 \pm 12}{10} = 2, \frac {22}5$

$q = -3p +7 = 1,\frac {-31}5$

• $p = 3q-10$ and $q = -3p+8$ don't satisfy the solutions given. – sku May 24 '18 at 21:37
• Then I'm made an arithmetic error. Do the same thing without making an error. – fleablood May 24 '18 at 22:07
• He's just pointing the error out. It is good practice to check your solution before posting I think – user211599 May 25 '18 at 0:06