How do you solve #(x-4)(x+3)<0#?

3 Answers
Jun 22, 2018

Answer:

#x in(-3,4)#
graph{(x-4)(x+3)<0 [-3.22, 4.576, -2.05, 1.85]}

Explanation:

we have
#(x-4)(x+3)<0# from here we get two critical points i.e X=4,-3
these are the values on which the sign of the equation changes
to get a range for x mark these critical points on a real number line
take any value between -3 and 4 (lets say 0)
#(0-4)(0+3)<0# always hence this part (-3,4) satisfies the equation
now take any value greater than 4 or less than -3

case 1(greater than 4)
#(5-4)(5+3)<0# ........is not valid as it gives a positive value on LHS
similarly
case -2(less than -3)
#(-6-4)(-6+3)<0#..........is also not valid as it gives a positive value on LHS
hence the required solution
#x in(-3,4)#

Jun 22, 2018

Answer:

You have to consider two possibilities:

Explanation:

For a product to be negative, one of the factors must be negative and the other positive.

(1)
#x-4<0andx+3>0#
This leads to:
#x<4andx> -3#, together this is #-3 < x<4#

(2)
#x-4>0andx+3<0#
This leads to:
#x>4andx<-3#
These two contradict.

So the solution space is #-3 < x<4#

Jun 30, 2018

Answer:

#-3 < x <4 #

Explanation:

This inequality is saying that the product of two things is less than zero, or in other words, negative.

If the product of two things is negative, one of them has to be negative. So we can say

#x-4<0# and #x+3>0#

We can't have the same signs, because that would make the solution set positive.

Isolating the #x# in both inequalities, we get

#x<4# and #x> -3#

These solutions overlap each other, as the only values of #x# that satisfy both are given by

#-3 < x <4 #

Hope this helps!