How do you solve and write the following in interval notation: #4x²+12x+9≤0#?

1 Answer
Jul 21, 2016

Answer:

#x=-3/2#

Explanation:

Note that as #4x^2+12x+9=(2x)^2+2xx2xxx3+3^2#, its factors using identity #(a+b)^2=a^2+2ab+b^2# are #(2x+3)^2#.

As #(2x+3)^2# is a perfect square, it cannot be negative and the only solution is

when #4x^2+12x+9=0# or #(2x+3)^2=0# i.e.

#2x+3=0# or #x=-3/2#

graph{4x^2+12x+9 [-2.402, 0.098, -0.28, 0.97]}