Exercise 137 Page 514

Practice makes perfect

a To solve for the boundary points we will treat the inequality as an equality and rearrange it so that the right-hand side equals

0 .

x^{2} + 2 x + 1 = 4

SubEqn

$LHS - 4 = RHS - 4$

x^{2} + 2 x - 3 = 0

If a second degree equation can be factored, we will be able to rewrite the

x -

term as a sum of the factors, and the constant as a product in the following way.

00 = = x^{2} x^{2} + + 2 (a + b) x x + + (- 3) a b

Let's list all of the ways that

3

can be factored, and investigate which pair of factors adds to

2 .

Notice that because the constant is negative one factor must be positive and the other must be negative.

integer 33 a b 1 (- 3) 3 (- 1) a + b - 2 2

By substituting

a = 3

and

b = - 1

into the equation

0 = (x + a) (x + b),

we will have factored it.

0 = (x + 3) (x + (- 1)) \Leftrightarrow 0 = (x + 3) (x - 1)

As we can see, the equation has two

x -

intercepts and therefore two boundary points.

b From Part A we know that there are two boundary points,

x = - 3

and

x = 1 .

Since there are two boundary points, we have a total of three surrounding areas.

All of these regions have to be tested.

c Let's test the three regions we identified from Part B.

$x$	$x^{2} + 2 x + 1$	$=$	$< 4 ?$
$- 4$	$(- 4)^{2} + 2 (- 4) + 1$	$7$	No
$- 1$	$(- 1)^{2} + 2 (- 1) + 1$	$0$	Yes
$2$	$2^{2} + 2 (2) + 1$	$9$	No

The results tells us that between $- 3$ and $1$ the inequality is true. Notice that the original inequality is strict, which means we have to exclude the boundary points from the solution set. With this information we can write the inequality and draw the solution set.