How do you solve #x^{2} + 3x + 6< 0#?

1 Answer
Sep 22, 2016

There are no Real values of #x# satisfying:

#x^2+3x+6 < 0#

In other words, the solution set is empty.

Explanation:

Method 1

Complete the square to find:

#x^2+3x+6 = (x^2+3x+9/4)-9/4+6#

#color(white)(x^2+3x+6) = (x+3/2)^2+15/4#

Note that for any Real value of #x#, we have:

#(x+3/2)^2 >= 0#

and hence:

#x^2+3x+6 = (x+3/2)^2 +15/4 >= 15/4#

So there is no Real value of #x# for which:

#x^2+3x+6 < 0#

Method 2

#x^2+3x+6# is in the form #ax^2+bx+c# with #a=1#, #b=3# and #c=6#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 3^2-4(1)(6) = 9-24 = -15#

Since #Delta < 0#, we can deduce that #x^2+3x+6# has no Real zeros.

Then since the coefficient of #x^2# is positive, we can deduce that:

#x^2+3x+6 > 0#

for all Real values of #x#.

Hence there are no Real solutions to #x^2+3x+6 < 0#