Consider the following inequality:
−
x
2
+
6x
−
8
≤
0
−
x
2
+
6x
−
8
≤
0
size 12{  x rSup { size 8{2} } +6x  8 <= 0} {}
(1)
There are a number of different ways to approach such a problem. The one that is stressed in the text is by graphing.
We begin by graphing the function
y=−x2+6x−8y=−x2+6x−8 size 12{y=  x rSup { size 8{2} } +6x  8} {}. We know how to graph this function by completing the square, but we’re going to take a shortcut—one that will not actually tell us what the vertex is, but that will give us what we need to know. The shortcut relies on finding only two facts about the quadratic function: what are its roots (the places it crosses the
xx size 12{x} {}axis), and which way does it open?
Of course we know three ways to find the roots of a quadratic equation: the easiest is always factoring when it works, as it does in this case.
−
x
2
+
6x
−
8
=
0
−
x
2
+
6x
−
8
=
0
size 12{  x rSup { size 8{2} } +6x  8=0} {}
(2)
x
−
4
−
x
+
2
=
0
x
−
4
−
x
+
2
=
0
size 12{ left (x  4 right ) left (  x+2 right )=0} {}
(3)
x
=
0,
x
=
2
x
=
0,
x
=
2
size 12{x=0,x=2} {}
(4)
Second, which way does the parabola open? Since it is a vertical parabola with a negative coefficient of the x2x2 size 12{x rSup { size 8{2} } } {} term, it opens down.
So the graph looks like this:
(Hey, if it’s that easy to graph a quadratic function, why did we spend all that time completing the square? Well, this method of graphing does not tell you the vertex. It tells us all we need to solve the quadratic inequality, but not everything about the graph. Oh. Darn. Anyway, back to our original problem.)
If that is the graph of
y=−x2+6x−8y=−x2+6x−8 size 12{y=  x rSup { size 8{2} } +6x  8} {}, then let us return to the original question: when is
−x2+6x−8−x2+6x−8 size 12{  x rSup { size 8{2} } +6x  8} {}less than or equal to 0? What this question is asking is: when does this graph dip below the
xx size 12{x} {}axis? Looking at the graph, the answer is clear: the graph is below the
xx size 12{x} {}axis, and therefore the function is negative, whenever
x≤2x≤2 size 12{x <= 2} {}orx≥4x≥4 size 12{x >= 4} {}.
When students graph these functions, they tend to get the right answer. Where students go wrong is by trying to take “shortcuts.” They find where the function equals zero, and then attempt to quickly find a solution based on that. To see where this goes wrong, consider the following:
x
2
−
2x
+
2
≥
0
x
2
−
2x
+
2
≥
0
size 12{x rSup { size 8{2} }  2x+2 >= 0} {}
(5)
If you attempt to find where this function equals zero—using, for instance, the quadratic formula—you will find that it never does. (Try it!) Many students will therefore quickly answer “no solution.” Quick, easy—and wrong. To see why, let’s try graphing it.
y
=
x
2
−
2x
+
2
=
x
2
−
2x
+
1
+
1
=
x
−
1
2
+
1
y
=
x
2
−
2x
+
2
=
x
2
−
2x
+
1
+
1
=
x
−
1
2
+
1
size 12{y=x rSup { size 8{2} }  2x+2=x rSup { size 8{2} }  2x+1+1= left (x  1 right ) rSup { size 8{2} } +1} {}
(6)
The original question asked: when is this function greater than or equal to zero? The graph makes it clear: everywhere. Any
xx size 12{x} {} value you plug into this function will yield a positive answer. So the solution is: “All real numbers.”
This technique is not stressed in this book. But it is possible to solve quadratic inequalities without graphing, if you can factor them.
Let’s return to the previous example:
−
x
2
+
6x
−
8
≤
0
−
x
2
+
6x
−
8
≤
0
size 12{  x rSup { size 8{2} } +6x  8 <= 0} {}
(7)
x
−
4
−
x
+
2
≤
0
x
−
4
−
x
+
2
≤
0
size 12{ left (x  4 right ) left (  x+2 right ) <= 0} {}
(8)When we solved quadratic equations by factoring, we asked the question: “How can you multiply two numbers and get 0?” (Answer: when one of them is 0.) Now we ask the question: “How can you multiply two numbers and get a negative answer?” Answer: when the two numbers have different signs. That is, the product will be negative if...
Table 1
The first is positive and the second negative 
OR 
The first is negative and the second positive 
x−4≥0x−4≥0 size 12{x  4 >= 0} {} AND
−x+2≤0−x+2≤0 size 12{  x+2 <= 0} {} 
OR 
x−4≤0x−4≤0 size 12{x  4 <= 0} {} AND
−x+2≥0−x+2≥0 size 12{  x+2 >= 0} {} 
x≥4x≥4 size 12{x >= 4} {} AND
−x≤−2−x≤−2 size 12{  x <=  2} {} 
OR 
x≤4x≤4 size 12{x <= 4} {} AND
−x≥−2−x≥−2 size 12{  x >=  2} {} 
x≥4x≥4 size 12{x >= 4} {} AND
x≥2x≥2 size 12{x >= 2} {} 
OR 
x≤4x≤4 size 12{x <= 4} {} AND
x≤2x≤2 size 12{x <= 2} {} 
x
≥
4
x
≥
4
size 12{x >= 4} {}

OR 
x
≤
2
x
≤
2
size 12{x <= 2} {}

This is, of course, the same answer we got by graphing.
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