Graphing with exponents

Question

I am given the following equation:

 

x² + 3x > 4, D = {Reals}

 

I am told to graph. How do you solve this equation?

Answer 1

Subtract 4 on both sides of inequality.

 

x² + 3x - 4 > 0

 

 

Factor the left side of inequality.

 

(x + 4)(x - 1) > 0

 

 

Now, we will set the factors equal to zero.  We do this because the x values that result from this will be on the x-axis.  After that, we will use a method to find the values that is above the x -axis.

 

x + 4 = 0                 and             x - 1 = 0

x = -4                       and            x = 1

 

 

Next, we will use test points around these x values and substitute them into the inequality.  We will use -5, -1 , and 2 as our test points.

 

 

f(-5) = (-5 + 4)(-5 - 1)

        = (-1)(-6)

        = 6

 

 

f(-1) = (-1 + 4)(-1 - 1)

      = (3)(-2)

      = -6

 

f(2) = (2 + 4)(2 - 1)

      = 6

 

The inequality is true when x=-5 and x=2.

 

 

The solution is

 

(-∞, -4)∪(1, ∞)

 

 

To graph the solutions

 

Draw a number line with the points -4 and 1.  Draw an open circle at -4 with an arrow pointing left from that circle.  Draw an open circle at 1 with an arrow pointing right from that circle.  In the end, there should be a gap between -4 and 1.

Answer 2

x² + 3x - 4 > 0         

(x + 4)(x - 1) > 0             for the product to be >0, both factors must have the same sign.

 

1. x + 4 > 0                and         x - 1 > 0

    x > -4                     and         x > 1               So one solution is x > 1

 

                        OR

 

2. x + 4 < 0                and         x - 1 < 0

    x < -4                     and         x < 1               So one solution is x < -4

 

Put these two on a number line.