Solve the inequality: x2 - 5 > 0?

Answer 1

Note: I'm assuming that x2 means x squared. Use x^2 for x squared (because technically x2 means x times 2 which is 2x).
First set x^2 - 5 = 0 and solve for x.
Add 5 to each side
x^2 = 5
Take the square root of each side (Don't forget that it could be + or -)
x = sqrt(5) or -sqrt(5)
Because you have two x-values you will have 2+1 = 3 intervals. You always have one more interval than you do values.
List the x-values in numerical order.
-sqrt(5), sqrt(5)
Your intervals are
(-∞, -sqrt(5)), (-sqrt(5), sqrt(5)) and (sqrt(5),∞)
Pick a point in each interval and test them in your original inequality. You are looking for the intervals that give you a positive number. You will see that any number in the interval (-sqrt(5), sqrt(5)) will give you a negative value.
Answer: (-∞, -sqrt(5)) U (sqrt(5),∞)
Note: ∞ is the symbol for infinity.
You could also write your answer as
x < -sqrt(5) or x > sqrt(5)

Answer 2

resolve the inequality: x2 - 5 > 0? in the beginning, the question could truly say 2x no longer x2. 2x - 5 > 0 Secondly, deliver 5 to the different part of the inequality. So, -5 turns into +5. Do you spot why? we've: 2x > 5 next: Divide the two factors of the inequality by 2 to unravel for x. very final answer: x > 5/2. What does our answer recommend? It potential that even if the fee of x is, it is extra effective than the fraction 5/2. i'm hoping this facilitates. Guido P.S. shop this thoughts: while multiplying or dividing by a damaging selection, you may substitute the process the sign on your very final answer.

Answer 3

The given equation does not admit of a solution whenever x is equal to 0; and:
x< -sqrt(5); when x is negative
x> sqrt(5); when x is positive

EXPLANATION:

x^2 - 5 > 0; gives |x| >sqrt(5), where |x| = Abs(x)
Considering the definition of the Abs function, the result follows immediately.

For a definition of Abs function, refer to the following location:
http://www.filemaker.com/help/FunctionsRef-318.html

Answer 4

x2 >5

x > + ó - square root 5

Answer 5

x2 - 5 > 0

x2 > 5
x>+sqrt5 or x>-sqrt5