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Solving Quadratic Inequalities: Examples (page 2 of 3)

• Solve 2x² + 4x >   x² – x – 6.

The two associated two-variable equations in this case are y = 2x² + 4x and y = x² – x – 6.

This inequality is asking when the parabola for y = 2x2 + 4x (in green) is higher than the parabola for y = x2 – x – 6 (in blue):

As you can see, it is hard to tell where the green line (y = 2x² + 4x) is above the blue line
(y = x² – x – 6). So, instead of trying to solve this inequality, it is simpler to deal with the following related inequality:

2x² + 4x > x² – x – 6
2x² + 4x – x² + x + 6 > 0
x² + 5x + 6 > 0

This last inequality is simpler to deal with because now all I have to do is find the zeroes of y = x² + 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 = x² + 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 "2x² + 4x > x² – x – 6" to get "x² + 5x + 6 > 0". The associated two-variable equation is y = x² + 5x + 6. First, I'll find the zeroes (that is, the x-intercepts):

x² + 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 = x² + 5x + 6 above the x-axis? Since y = x² + 5x + 6 graphs as a right-side-up parabola, it is above the axis on the ends:

Then the solution is:

x < –3  or  x > –2

Why was this solution "or equal to", rather than just "greater than" or "less than"? Because the original inequality was "or equal to", so the boundary points, the zeroes (the x-intercepts), are included in the solution.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Solve x² + x + 1 > 0.

First, I'll find the zeroes:

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

Since y = x2 + 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 = x² + 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 x² + 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 = x² + 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, 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.

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