Restrictions of a Radical Example 6

Example 6 Problem and Work

f(x) = √(x2 - 4x - 5)

Example 6:  Algebraic Solution

x2 -4x - 5 = 0                  Set the radicand equal to 0

(x - 5) (x + 1) = 0               Factor

x - 5 = 0 x + 1 = 0             Set factors equal to 0

x = 5 x = -1                     The radicand should be ≥ 0

The Zeros are x = 5, -1

We will need to find the positive values.

    -2             0                6             These are the test points

<-------1------------5---------->

Plug in the test points in the radicand to determine if positive or negative.

x = -2            (-2 - 5) (-2 + 1)   =    (-7) (-1)     =>    positive

x = 0              (0 - 5) (0 + 1)     =     (-5) (1)      =>    negative

x = 6              (6 - 5) (6 + 1)     =     (1) (7)       =>    positive

    -2            0                 6

<+++++++++1]------------------ [5+++++++++++>

Solution set: (-inf, -1] U [5,inf)