What is the solution set for #-x^2 + 2x > -3#?

1 Answer
Aug 31, 2015

Answer:

#x in (-1,3)#

Explanation:

Start by getting all the terms on one side of the inequality. You can do that by adding #3# to both sides

#-x^2 + 2x + 3 > - color(red)(cancel(color(black)(3))) + color(red)(cancel(color(black)(3)))#

#-x^2 + 2x + 3 > 0#

Next, make the quadratic equal to zero in order to find its roots. This will help you factor it. Use the quadratic formula to calculate #x_(1,2)#.

#-x^2 + 2x + 3 =0#

#x_(1,2) = (-2 +- sqrt(2^2 - 4 * (-1) * (3)))/(2 * (-1))#

#x_(1,2) = (-2 +- sqrt(16))/((-2))#

#x_(1,2) = (-2 +- 4)/((-2)) = {(x_1 = (-2-4)/((-2)) =3), (x_2 = (-2 + 4)/((-2)) = -1) :}#

This means that you can rewrite the quadratic as

#-(x-3)(x+1)=0#

Your inequality will be equivalent to

#-(x-3)(x+1) > 0#

In order for this inequality to be true, you need one of the two terms to be positive and the other negative, or vice versa.

Your first two conditions will be

#x-3 > 0 implies x > 3#

and

#x + 1 < 0 implies x < -1#

Since you can't have values of #x# that are both greater than #3# and smaller than #(-1)#, this possibility is eliminated.

The other conditions wll be

#x - 3 < 0 implies x < 3#

and

#x + 1 > 0 implies x > -1#

This time, these two intervals will produce a valid solution set. For any value of #x# that is greater than #(-1)# and smaller than #3#, this product

#(x-3) * (x+1) <0#

which means that

#-(x-3)(x+1) > 0#

The solution set for this inequality will thus be #x in (-1,3)#.