We’ve talked before about solving systems of linear inequalities. There, we simply graph the relevant equations to find where the shaded regions all intersect. We do the same thing for systems of quadratic inequalities. Let’s look at an example:
We begin by graphing the two equations:
Then, we shade in the regions where each inequality applies. So, for example, where will ? Clearly this will be true in the region above the curve. (And if you are unsure, you can always just try testing a point on each side of the curve). We do the same for the second equation to get:
And as before, the solutions will be any point in the area where all of the shaded regions intersect.
Now, we can also have systems of inequalities that combine linear and quadratic inequalities:
We follow the same method, graphing the functions:
Shading the relevant regions:
And identifying where shaded regions intersect:
And that’s how we handle systems of inequalities for quadratic equations.
Practice Problems
For each of the following systems of quadratic inequalities, graph their solutions. (Some may not have any solutions at all).

Answer
It may not be obvious from this graph that there is no solution, but we can see that the only point which is apt to intersect both regions is the point (0, 0). Plugging in an xvalue of 0 and a yvalue of 0 reveals, though, that the point satisfies neither inequality.Another way to realize that this pair of equations has no solution would be to reason as follows:
forces
to be positive (since
for any value of
). But
means that
must be negative.
cannot be both positive and negative, so it cannot satisfy both inequalities at the same time. Since no value of
will do, this pair of inequalities has no solution.

Answer
It may be somewhat hard to tell how the different regions interact, but a careful inspection reveals that the four regions do not all overlap anywhere, which means this set of equations has no solution.Another way to see that they have no solution would be to reason as follows: note that the last inequality requires
. So for any solution,
. And since
, we conclude that
as well (since
implies
). But another one of our inequalities requires
! We cannot have
be greater than 5 and less than 5. So no value of
will work. So there is no solution.

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