Solve (x-2)(x+1)0

Question

Solve (x-2)(x+1)>0

anonymous · Answer

x>2 and x>-1

anonymous · Answer

x<-1 ,x>2

anonymous · Answer

wait why less than -1?

a_clan · Answer

If the product of two terms is greater than 0, it means either both terms are +ve or both _ve

anonymous · Answer

+ve?-ve?

anonymous · Answer

postive and negative

myininaya · Answer

(x-2)(x+1)=0 when x=2 or x=-1


so we have


-------|---------|--------
         -1                 2

we need to test these three intervals 

(-2-2)=-     (0-2)=-       (3-2)=+
(-2+1)=-     (0+1)=+      (3+1)=+
product is +  product is -   product is +

the answer is (-inf,-1) U (2,inf)

anonymous · Answer

attachment

a_clan · Answer

Either both (x-2)>0 and (x+1)>0 OR (x-2)<0 and (x+1)<0

anonymous · Answer

myininaya · Answer

what zak and i have are same
just different notation

anonymous · Answer

look 0>-1 if u put zero the inequility is not satisfied

anonymous · Answer

but it's x.. not 0?

anonymous · Answer

I subtracted the 1 over to the right and got x > -1

myininaya · Answer

did you not understood what i did mybrain?

anonymous · Answer

No I didn't, I'm sorry >.<

a_clan · Answer

If the product of two terms is greater than 0, it means either both terms are +ve or both _ve Either both (x-2)>0 and (x+1)>0 ........(1) OR (x-2)<0 and (x+1)<0 ........................(2) On solving (1), we get x>2 and x>-1 So, resultant solution will be x>2 which will satisfy both these above inequality. On solving, (2), we get x<2 and x<-1 So, solution x<-1 On combining the two solutions, (-infty,-1) U (2, infty)

myininaya · Answer

i found the zeros and i test the areas around the zeros to see if the function was above the x-axis or below

anonymous · Answer

OH i see

myininaya · Answer

when is (x-2)(x+1) , 0?

anonymous · Answer

now I understand why you did the "test" earlier.

myininaya · Answer

ok i just plug in a number before and after each of the zeros
to see if got a positive number or a negative number
and we want it to be positive since we have > and not <

anonymous · Answer

wait. if x < -1.. you could plug in say -3 and you would get -2.. which doesn't satisfy the equation x > 0...?

anonymous · Answer

Thank a lot, guys :) Now imagine, that defines set B (B = {x: (x-2)(x+1)>0}), and set A is defined by A = {x: x<3.5}, and i have to find the values of x which defines set A (intersection) B ?