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You have just learned that two quadratics can intersect each other at 0, 1, 2, 3 or 4 points.

This means that a quadratic-quadratic system of equations can have 0, 1, 2, 3 or 4 solutions.

Like linear-quadratic systems of equations, the solutions of a quadratic-quadratic system of equations can be found graphically or algebraically.

Let's look at an example.

EXAMPLE 1:
Solve the following quadratic-quadratic system by graphing: 1. x2 + y2 - 9 = 0 and 2. 9x2 - 4y2 - 36 = 0
The equation x2 + y2 - 9 = 0 defines a circle centred at the origin with a radius of 3. The equation 9x2 - 4y2 - 36 = 0 defines a hyperbola centred at the origin, opening left and right, with asymptotes of slope +3/2 and -3/2.
Sketching the graphs of both equations can give you a good estimate of where they intersect. The graphs are drawn for you in Figure 1.	FIGURE 1
Notice that the circle and the hyperbola intersect at 4 places. If you zoom in, you can estimate the points of intersection as: (2.35, 1.86), (-2.35, 1.86), (2.35, -1.86) and (-2.35, -1.86).

Like linear-quadratic systems of equations, the solutions of quadratic-quadratic systems can also be found algebraically.

Click to review the strategies for solving systems of equations algebraically.

For our example of this page, a good strategy (but not the only strategy) for solving the system would be the elimination method:

Check out these challenging questions that will help you practice solving quadratic-quadratic systems of equations.