Solve the quadratic inequality. Write the solution set in interval notation.

Question

x² - 8x + 15 > 0

Answer 1

You should start by solving the quadratic equation x² - 8x + 15 = 0. This can be done by factoring or the quadratic formula, as necessary.

 

(x-5)(x-3) = 0

 

x - 5 = 0 OR x - 3 = 0

 

x = 5 OR x = 3

 

Next, mark the two numbers 3 and 5 on the number line. (I cannot show this here now.) There are three intervals formed:

 

(-∞, 3) , (3,5), and (5,∞).

 

What you must do next is choose a number inside each of these intervals, plug it into the ORIGINAL inequality, and see if it becomes true or false. If the chosen number makes it true, then ALL numbers in that interval will solve the problem; if the chosen number makes it false, then NONE of the numbers in that interval will be a solution.

 

Here we go:

 

x = 0 → 15 > 0 which is true → (-∞, 3) is in the solution

x = 4 → 4² - 8*4 + 15 = 16 - 32 + 15 = -1 > 0 is false → (3,5) is not part of the solution

x = 6 → 6² - 8*6 + 15 = 36 - 48 + 15 = 3 > 0 is true → (5,∞) is in the solution

 

All together, the solution is written in interval notation as:

 

(-∞,3) ∪ (5,∞).

 

If you need to graph the solution as well, mark a circle at 3 and 5 on a number line, then shade to the ends of the line (do not shade between 3 and 5). I cannot show that here now.

Answer 2

x² - 8x + 15 > 0

(x-3)(x-5) > 0        [this is the same inequality]

 

This is true for:

 (x-3)   (x-5)

  -----   -----

   +        +

    -         -

and none other.

 

So, either   (x-3)>0 AND (x-5)>0

                   (x>3) AND (x>5)

                          (x>5)

 

       or       (x-3)<0 AND (x-5)<0

                    (x<3) AND (x<5)

                            (x<3)

 

Thus, either  (x<3) or (x>5).

In interval notation,  (-∞,3) ∪ (5,∞)