Totally lost, please HELP with this question?

Question

Solve for x: x^2 + 3x – 10 > 0, x is a set of real numbers.

a) x > 2 or x < -5
b) x > - 2 or x < -5
c) x > -5 or x >2
d) x > -5 or x <2

Answer 1

x^2 _ 3x - 10 = (x+5)(x-2)

This is >0 if both are positive or both are negative.

To both be positive, x must be > 2. See what makes both be negative and you have it.

Answer 2

If you factor the left side of the equation you will get the values for x where the equation is equal to zero. This will determine 3 intervals where you can check whether the function is greater than or less than zero. Since they only give you answers of 2, 5 it's a pretty good guess that it will be easily factored. Pick a point in each interval to decide if the points on that interval make the statement true.

Answer 3

I'm not 100% sure, but as I read it..(x-2)(x+5)>0 for the expressiion to be greater than zero then x has to be greater than 2. OR x must be less than -5. So you have either 2 negatives equal more than zero or to positives equal more than zero.

Answer 4

First you factor to get: (x^2 + 5x) + (-2x- 10)
Then you factor again: x(x+5) - 2(x + 5)
Rewrite it as: (x - 2)(x + 5) > 0
Add 2 and subtract 5 from x and you get:
c) x> -5 or x> 2

Answer 5

First factor (x+5)(x-2)>0 so x>-5 and x>2
Inequalities should be checked after you find solutions by trying test solutions.

Answer 6

(x + 5).(x - 2) > 0
Answer c)

Answer 7

from my calculations, the first option is right

Answer 8

c