How do you solve #4x ^ { 2} + 48x + 108\geq 0#?

1 Answer
Binayaka C.
Dec 4, 2016

Solution: #x<=-9# or #x>=-3#. In interval notation #(-oo,-9]uu[-3,oo)#.

Explanation:

#4x^2+48x+108>=0 or x^2+12x+27>=0 or (x+9)(x+3)=0#
Critical points are #x+9=0 :.x=-9# or #x+3=0 :.x=-3#. From graph it is revealed that #y>=0# when #x<=-9# or #x>=-3#.
Solution: #x<=-9# or #x>=-3#. In interval notation #(-oo,-9]uu[-3,oo)#. graph{x^2+12x+27 [-20, 20, -10, 10]}[Ans]

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