Solving
Quadratic Inequalities: Concepts (page
1 of 3)

Solving linear inequalities,
such as "x
+ 3 > 0", was
pretty straightforward, as long as you remembered to flip the inequality
sign whenever you multiplied or divided through by a negative (as you
would when solving something like "–2x
< 4").

There is a big jump, though,
between linear inequalities and quadratic inequalities. Part of the jump
is the fact that concepts which were skipped over in learning how to solve
linear inequalities are useful, even needful, in solving quadratic inequalities.
So let's first look at a linear inequality, and cover those concepts that
were skipped earlier.

Solve x
– 4 < 0.

I already know that,
to solve this inequality, all I have to do is add the 4
to the other side to get the solution "x
< 4". So
I already know what the answer is. But now I'll approach this problem
from a different angle, by considering the related two-variable graph.

For "x
– 4 < 0",
the associated two-variable linear graph is y
= x – 4:

The inequality "x
– 4 < 0"
is asking "when is the line y
= x – 4 below
the line y
= 0?" Since
the line y
= 0 is just the x-axis,
the inequality is therefore asking "when is the line y
= x – 4 below
the x-axis?"
The first step in answering this question is to find where the line
crosses the x-axis;
that is, first I need to find the x-intercept.
So I set y
equal to zero and solve:

Since
the slope
of this line is m
= 1 (in particular,
since the slope is positive), then the line is increasing, so
the line is below the axis on the left-hand side (before the intercept)
and above the axis on the right-hand side (after the intercept),
as is highlighted at right:

The original question
asked me to solve x
– 4 < 0,
so I need to find where the line is below the x-axis.
This happens on the left-hand side of the intercept:

Since the original
inequality, "x
– 4 < 0",
asked only about the x-values,
I'll restrict the above graph to just the x-axis:

Thinking back to the
graphical method of presenting solutions to linear
inequalities,
the above graph displays the correct solution "x
< 4".

That is, by looking at
the graph of the associated line and determining where (on the x-axis)
the graphed line was below the x-axis,
you can easily see that the solution to the inequality "x
– 4 < 0" is
the inequality "x
< 4". You can
follow the same method of finding intercepts and using graphs to solve
inequalities containing quadratics.

Let's look at a quadratic
inequality:

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Solve –x^{2}
+ 4 < 0.

First, I need to look
at the associated two-variable equation, y
= –x^{2} + 4,
and consider where its graph is below the x-axis.
To do this, I need to know where the graph crosses the x-axis.
That is, I first need to find where –x^{2}
+ 4 is equal to zero:

–x^{2}
+ 4 = 0
x^{2}
– 4 = 0
(x
+ 2)(x – 2) = 0
x
= –2 or
x
= 2

This says that the quadratic
crosses the x-axis
at x
= –2 and at x
= 2.

These zeroes
divide the number line into three intervals:

Now I need to figure
out where (that is, on which intervals) the graph is below the axis.
But that's easy! Since this is a "negative" quadratic, it
graphs
as an upside-down parabola.

In other words,
the graph is high (above the axis) in the middle, and low (below
the axis) on the ends:

To solve the original
inequality, I need to find the intervals where the graph is below
the axis (so the y-values
are less than zero).

My knowledge of
graphing, together with the zeroes I found above, tells me that
that I want the intervals on either end, rather than the interval
in the middle:

Then the solution is
clearly:

x
< –2 or x
> 2

I could have multiplied
the initial inequality through by –1,
giving me "x^{2}
– 4 > 0". The
zeroes would have been the same: x
= –2 and x
= 2. But this parabola
would have been right-side-up, since the quadratic would have been "positive".
That's okay though, because, by multiplying through by –1,
I would have flipped the inquality, so I would have been looking for where
the quadratic is greater than zero (that is, where the parabola
is above the axis). Since the parabola would have been right-side-up,
the graph would have been above the axes on the ends; so the solution
would have worked out to be the same as before: x
< –2 or x
> 2: