Solve the inequality x^2 - x - 12 > 0.?

Question

Options are:

A. ( -ve infinity , -4 ) U ( 3 , +ve infinity )

B. x = 4 , x = -3

C. ( -3 , 4 )

D. ( -ve infinity , -3 ) U ( 4 , +ve infinity )

Please try to give a step by step solution...

Answer 1

The answer is D.  The way you solve this is by finding the factors.  In this case,     (x-4)(x+3)>0 x could be > 4 or <-3.  But since this is an inequality, you need to include all possible answers. In this case, x could be any number except 3, 2, 1, 0, -1, or -2.  If you don't believe me, try plugging them into the equation.  It doesn't work.  

The hardest part of these problems is finding the factors.  You just have to find 2 numbers that multiply to the last number (in this case,-12) and add up to the middle number (-1x). The only possibility is -4 and +3.  So,   (x-4) and (x+3) are both > 0.  Then, you take both of those problems and solve for x.  x>4 and x>-3.

By plugging numbers into the inequality, you can see that x can be any number except those that fall in between 4 and -3.

Hope this helps.  It looks kind of sloppy and I haven't done this in a looooonnnnng time.

Answer 2

x^2 - x - 12 > 0

(x-4)(x+3)>0

from the zero property x=4 and x = -3

now check values for x aroudn the above numbers

u shoudl find its D

Answer 3

(x - 4)(x + 3) > 0

x - 4 > 0 AND x + 3 > 0

x > 4 and x > - 3

x > 4

x - 4 < 0 , x + 3 < 0

x < 4 , x < - 3

x < - 3

{ x : x < - 3 U x > 4}    

OPTION D    

Answer 4

x^2 - 4x + 3x -12 >0

=> x(x-4)+3(x-4)>0

=>(x+3)(x-4)>0

When x is greater than 4 both x+3 and x-4 is greater than 0

When x is less than -3 both x+3 and x-4 is less than 0

Therefore the answer is D.