solve for x where x is a real number... x^2 - 2x - 15

Question

solve for x where x is a real number... x^2 - 2x - 15 < = 0

anonymous · Answer

$$x^2 - 2x - 15 \le 0$$

anonymous · Answer

Solve for x over the real numbers:
x^2-2 x-15 = 0
Factor the left hand side.
The left hand side factors into a product with two terms:
(x-5) (x+3) = 0
Solve each term in the product separately.
Split into two equations:
x-5 = 0 or x+3 = 0
Look at the first equation:  Solve for x.
Add 5 to both sides:
x = 5 or x+3 = 0
Look at the second equation:  Solve for x.
Subtract 3 from both sides:
Answer: |  
 | x = 5 or x = -3

anonymous · Answer

@zkrup what if it's $$x^4 -1 / x^3 = 0$$

anonymous · Answer

how would I do that? @zkrup