# Quadratic Inequalities

An inequality expresses the relation between two
expressions where one is greater than the other.

Unlike equations , inequalities typically have an infinite number of solutions
(and an infinite number of

non-solutions).

We solved inequalities in elementary algebra using the
techniques for solving equations:

EX: Solve:

Note that when dividing or multiplying by a negative we
must reverse the inequality symbol since

"largeness" reverses when we move from the negative to the positive side of the
number line. For

example, 5 > 3 but -5 < -3.

We can show the infinite number of solutions on a number line.

Note the open circle at -2 to indicate that -2 is NOT part
of the solution.

We can write the solution as follows:

-2 > x

or x < -2

or {x | x < -2}

or (-∞,-2)

We can check the solution by replacing x with a value that
is in the solution set and seeing if it

satisfies the inequality. We can also choose a "non-solution" and see if that
does NOT work.

We can also think of

3 + 2x < -3x - 7 in terms of functions .

For example:

Let f (x) = 3 + 2x and g (x) = -3x - 7

Solve f (x) < g (x) .

That is, find the values of x where f < g.

f and g are two lines that we can easily graph:

We want the values of x where the blue g is above (greater than) the red f.

We can see that g is above f when x is less than –2.

That is, the solution to f (x) < g (x) is {x | x < -2} .

EX: Consider the function f (x) = x^{2} - 6x + 5 . Solve the
inequality f(x) < 0

We can graph the function and locate the region where the graph is below the
x-axis.

vertex:

x-int:

y-int:

Since we want
f (x) < 0 , we are looking for the values of x where f is below the x-axis.

Thus, the solution is {x |1< x < 5}.

Let's check 3 values:

EX: Solve the inequality: -x^{2} < 4x -12

We can solve this in several different ways .

** Method A : **Move all terms to one side of the inequality:

Now, graph the functionf (x) = -x^{2} - 4x +12

The vertex is (-2, 16), the x-intercepts are (-6, 0) and

(2, 0), and the y-intercept is (0, 12).

On the graph, note where the function is below the x-axis:

So, the solution is {x | x < !6 or x > 2}

Note that if we rearrange the inequality into an
equivalent

inequality we will get a different graph but the same solution.

For example:

This is an upward opening parabola with vertex (-2, -16),

x-intercepts (-6, 0) and (2, 0), and y-intercept (0, -12).

Here, we want to locate the values of x where the function is

**positive** rather than **negative**. The solution is the same as

before, {x | x < -6 or x > 2} .

**Method B: **We can also solve -x^{2} < 4x - 12 by defining two
functions and then finding the values of

x where one is greater than the other.

Let f (x) = -x^{2} and g (x) = 4x - 12 .

We have a standard downward opening parabola and a line

of slope 4 and y -intercept -12.

How, we want to know for which values of x we have

f (x) < g (x) . On the graph, these are the values of x where

the red f(x) is below the blue g(x).

We get the same solution as before.

**Method C:** We can also solve -x^{2} < 4x - 12 algebraically,
without graphing.

**Step 1: **Rearrange the inequality: 0 < x^{2} + 4x - 12 Now, we want to know for
which values of x is

0 < f (x). In other words, when is this function positive?

**Step 2:** Factor the expression:

f (x) = x^{2} + 4x - 12

= (x - 2)(x + 6)

We want to know when this product is positive ; that is, when

0 < f (x). That will be the case when

the signs of the two factors are the same (both negative or both positive).

**Step 3:** We know that the only way a factor can switch sign is when it passes
through 0. So, separate

the number line into intervals using values where the factors are 0:

**Step 4:** Determine the region where the function obeys the
inequality,
0 < f (x).

We can see that the function is positive in the intervals x <
-6 and x > 3.

We will use this technique when solving inequalities involving rational
polynomials in section 4.4

p157 #36

John Deere found that the revenue of tractors is given by R(p) = -(1/2)p^{2} +1900p
where p is the unit

price in dollars. For what range of prices will revenue exceed 1,200,000?

Let R(p) = -(1/2)p^{2} +1900p and g (p) = 1,200,000. We want
to know for what values of p is

R(p) > g (p) .

From the graph we can estimate the solution as:

{p | 800 < p < 3000}.

**EX: Solve **25x^{2}+16 < 40x

Let's solve this algebraically:

**Step 1: **Rearrange the inequality: 25x^{2}- 40x +16 < 0 Now, we want to know for
which values of x is

f (x) < 0 . In other words, when is this function negative?

**Step 2:** Factor the function:

25x^{2}- 40x +16 < 0

(5x - 4)(5x - 4) < 0

We want to know when this product is negative. That will be the case when the
signs of the two

factors are different.

**Step 3:** Separate the number line into intervals using values where the factors
are 0. The zeros of the

factors are

**Step 4:** Determine the region where the function obeys the
inequality,

f (x) < 0 . The function is never

negative so the answer is no solution.

Note that we could have seen this by writing (5x - 4)(5x - 4) < 0 as (5x - 4)^{2}<
0 .

The square of a real number cannot be negative so right here we can see that
there is no solution.

Let's check this using graphs:

Let f (x) = 25x^{2}+16 and g (x) = 40x .

We want the values of x where f (x) < g (x) :

We see that g is never above f, although it

appears that g and f touch at about x = 4/5.

**EX: Solve** x(x +1) > 20

Let's solve this algebraically:

**Step 1: **Rearrange the inequality: x^{2} + x - 20 > 0 Now, we want to know for
which values of x is

f (x) > 0 . In other words, when is this function negative?

**Step 2:** Factor the function:

x^{2} + x - 20 > 0

(x - 4)(x + 5) > 0

We want to know when this product is negative. That will be the case when the
signs of the two

factors are different.

**Step 3:** Separate the number line into intervals using values where the factors
are 0 as cutoffs. The

zeros of the factors are at x = 4 and x = -5.

**Step 4:** Determine the region where the function obeys the inequality,f (x) > 0 .

This is {x | -5 < x or x > 4}

Let's check this using graphs:

Let f (x) = x(x +1) and g (x) = 20 .

We want the values of x where

f (x) > g (x) :

We see that f is above g when

x < -5 or when x > 4.

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