[Common Math 1 Part 18] Systems of Quadratic Equations

In the previous post, we studied cubic and quartic equations by reducing them to lower-degree equations. Now we study systems of quadratic equations: finding ordered pairs that satisfy two equations at the same time.

A system of quadratic equations asks for the intersection points of two graphs. Algebraically, we usually substitute or eliminate to reduce the system to one equation.

1. What does a solution mean?

For

$$\begin{cases} y=x^2-2x\\ y=x+2 \end{cases}$$

a solution is an ordered pair ((x,y)) satisfying both equations. Graphically, it is an intersection point of the parabola and the line.

2. If one equation is linear, substitute

Since both right sides equal y,

$$x^2-2x=x+2$$

so

$$x^2-3x-2=0.$$

Thus

$$x=\frac{3\pm\sqrt{17}}{2}.$$

Substitute into (y=x+2) to get

$$\left(\frac{3+\sqrt{17}}{2},\frac{7+\sqrt{17}}{2}\right), \quad \left(\frac{3-\sqrt{17}}{2},\frac{7-\sqrt{17}}{2}\right).$$

3. If both equations are quadratic, reduce when possible

For

$$\begin{cases} y=x^2-4x+1\\ y=-x^2+2x+7 \end{cases}$$

set the right sides equal:

$$x^2-4x+1=-x^2+2x+7.$$

Then

$$x^2-3x-3=0,$$

so

$$x=\frac{3\pm\sqrt{21}}{2}.$$

Substitute each value of x into either original equation to find y.

4. Sometimes adding or subtracting is useful

For

$$\begin{cases} x^2+y^2=10\\ x^2-y^2=2 \end{cases}$$

adding gives (2x^2=12), so (x^2=6). Subtracting gives (2y^2=8), so (y^2=4). Therefore the solutions are

$$(\sqrt6,2),\ (\sqrt6,-2),\ (-\sqrt6,2),\ (-\sqrt6,-2).$$

5. Common mistakes

• Do not stop after finding x; the answer is usually an ordered pair.
• Substitute back to find y.
• Check the ordered pairs in the original equations.
• Choose substitution or elimination depending on the form of the system.

6. Practice

7. Summary

A system of quadratic equations is a problem of finding graph intersections by algebra.