Solving
Quadratic Inequalities: Examples (page
2 of 3)

Solve2x^{2}
+ 4x>x^{2} – x
– 6.

The two associated two-variable
equations in this case are y
= 2x^{2} + 4x
and y
= x^{2} – x – 6.

This inequality
is asking when the parabola for y
= 2x^{2} + 4x
(in green) is higher than the parabola for y
= x^{2} – x – 6
(in blue):

As you can see, it is
hard to tell where the green line (y
= 2x^{2} + 4x)
is above the blue line (y
= x^{2} – x – 6).
So, instead of trying to solve this inequality, I will instead work
with the following related inequality:

This last inequality
is simpler to deal with because now all I have to do is find the zeroes
of y
= x^{2} + 5x + 6
(which is easy) and then pick the correct intervals based on just the
one parabola (which is also easy). That is, it is simpler to compare
one parabola with the x-axis
than to compare two parabolas with each other. But since the one parabola
(y
= x^{2} + 5x + 6)
came from combining the two original parabolas ("paraboli"?),
the solution to the simpler one-parabola inequality will be the same
as the solution to the original two-parabola inequality. Since the solutions
will be the same, I'll work with the simpler case.

I have simplified "2x^{2}
+ 4x>x^{2} – x – 6"
to get "x^{2}
+ 5x + 6 > 0".
The associated two-variable equation is y
= x^{2} + 5x + 6.
First, I'll find the zeroes (that is, the x-intercepts):

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x^{2}
+ 5x + 6 = 0
(x
+ 2)(x + 3) = 0
x
= –2 or
x
= –3

These two intercepts
split the number-line into three intervals, namely x
< –3, –3 < x < –2,
and x
> –2. On which
of these three intervals is y
= x^{2} + 5x + 6
above the x-axis?
Since y
= x^{2} + 5x + 6
graphs as a right-side-up parabola, the quadratic is above the axis
on the ends:

Hmm...
Since there is a negative inside the square root, there must not be
any x-intercepts.
That is, this quadratic must be either always above the x-axis
or else always below, because it can never cross or touch the axis.

Since
y
= x^{2} + x + 1
is a "positive" quadratic, the parabola is right-side-up,
so I know it goes up forever. For the parabola not to cross the
x-axis,
it must be that the parabola is always above the axis, as you
can see here:

So
when is y
= x^{2} + x + 1
greater than zero (above the axis)? Always! Then the solution is:

all
x

The
above solution could also be stated as "all real numbers" or
written as the interval "from negative infinity to positive infinity".

Solve
x^{2} +
x + 1 < 0.

This
looks just like the previous problem, except that now I'm looking for
where the parabola is below the axis. I already know that there are
no x-intercepts.
Also, because this is a right-side-up parabola, I know that the graph
is always above the axis. So where is y
= x^{2} + x + 1
less than zero? Nowhere! Then the solution is:

no
x

The
above solution could also be stated as "no solution" or as "the
empty set", represented by the character "Ø".

Whenever
you have a quadratic inequality where the associated quadratic equation
does not have real solutions (that is, where the associated parabola does
not cross the x-axis),
the solution to the inequality will either be "all x"
or "no x",
depending upon whether the parabola is on the side of the axis that you
need.