How do you solve #x^{2} + 3x + 6< 0#?
1 Answer
There are no Real values of
#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+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^2+3x+6 < 0#
Method 2
This has discriminant
#Delta = b^2-4ac = 3^2-4(1)(6) = 9-24 = -15#
Since
Then since the coefficient of
#x^2+3x+6 > 0#
for all Real values of
Hence there are no Real solutions to