Nonlinear Inequalities and the Pythagorean Theorem
Quadratic Inequalities
We will solve using the following steps

Put everything on the
left hand side so that we have for example
Quad > 0.

Factor and set equal
to
zero.

Solve and place answers on a number
line. This will cut the
number line into two or three regions.

Pick a test value for each region and plug that test value into
each of the factors. Put plusses or minuses over the region depending
on whether the test values test positive or negative.

If the region has two plusses or two minuses the the region is
positive. If the region has one of each then the region is
negative.

If the inequality is "<" then
include the
negative regions.
If the inequality is ">" then include the
positive regions.

If the inequality is a less (greater) than or equal to
then include the
endpoints with solid dot and the interval
[]
If the inequality is a less (greater) than then do not include the endpoints
by showing an open dot and the interval ()
Example
Solve
x^{2} + 3x
> 2

x^{2} + 3x + 2 > 0

(x + 2)(x + 1) = 0

x = 2 or
x = 1 This cuts the number line into
three regions.

For the left region we choose 5 and have that
5 + 2 < 0,
5 + 1 < 0
(two negatives)

For the middle region we choose 1.5 and have that
1.5 + 2 > 0,
1.5 + 1 < 0 (one of each)
For the left region we choose 0 and have that
0 + 2 >0, 0+ 1 >0
(two positives)

x+2 
x+1 
Total 
Left (5) 
 
 
+ 
Middle(1.5) 
 
+ 
 
Right(0) 
+ 
+ 
+ 

We see that the left and right regions are positive and the middle
region is negative.

Since the inequality is a greater then we have the solution:
(
,2) U (1,)
Exercise
Solve
x^{2}  3
< 2x
Rational Inequalities
To solve rational inequalities, we can use the same technique that we used
for quadratic inequalities with the following adjustments:

After putting everything on the left hand side we put the left hand
side over a common denominator.

Instead of factoring and solving to find the cut points, we just
set the numerator and denominator each equal to
zero.

The cut point that is determined from that
denominator will
never
be included, thus will be bordered by a "(" and shown as an open dot.
Example
2x + 3
< 2
x 1
2x + 3
 2 <
0
x 1
2x +
3 2(x  1)

< 0
x
1
x  1
2x + 3  2x +
2
< 0 Notice
that  2(x  1) = (2x  2) = 2x + 2
x 1
5
< 0
x 1
Notice that the numerator is never zero and the denominator is zero
only at x = 1.
x = 1 cuts the number line into two regions.

For the left region, we choose 0 and see that
5 > 0 and
0  1 < 0 (one of each)
For the right region we choose 2 and see that
5 > 0 and 2  1 > 0 (two positives)

x1 
Total 
Left (0) 
 
 
Right(2) 
+ 
+ 

We see that the left region is negative and the right region is
positive.

Since the inequality is "<" we include that left region.

Although the inequality as an equality under it, the cut point comes from the
denominator, therefore we do not include that cut point. Hence the
solution is
(,1)
The Pythagorean Theorem
The most important formula in mathematics is the Pythagorean
theorem which
states that for a right triangle, if a and b are the lengths of the legs
(short sides) and c is the length of the hypotenuse
(long side), then
a^{2} + b^{2} = c^{2}
Example
Suppose the length of hypotenuse of a right triangle has length 12 and a
leg has length 9. Find the length of the other leg.
Solution
Let x be the length of the other leg, then
x^{2} + 9^{2}
= 12^{2}
or
x^{2} + 81
= 144
or
x^{2} = 63
so that
x =
The length
of the other leg is
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