How do you solve the inequality #x^2+3x-28<0#?

1 Answer
Apr 4, 2018

Answer:

#-7 < x < +4#

Explanation:

#x^2+3x - 28 = x^2+7x-4x-48 = x(x+7)-4(x+7) = (x-4)(x+7)#

So, the inequality is

#(x-4)(x+7)<0#

Now. the left hand side vanishes for #x = -7# and #x=+4#.

If #x<-7#, then both the factors #(x+7)# and #(x-4)# are negative, and the product is positive. On the other hand, #if x> +4#, both factors are positive, leading once again to a positive product.

Thus the inequality is only satisfied when

#-7 < x < +4#

where the facor #(x+7)# is positive and #(x-4)# is negative.

You can see this in the graph below

graph{x^2+3x-28 [-10, 10, -40, 40]}