How do you solve the inequality #2x^2+9x+3<=0#?

3 Answers
Dec 22, 2016

Answer:

#x=frac{-9+-sqrt57}{4}#

#x approx -0.362541#
#x approx -4.1374586#

Explanation:

graph{2x^2+9x+3 [-10, 10, -5, 5]}

I personally like to visualize the inequality, and think "where is the graph of #2x^2+9x+3# less than or equal to zero". So we can set the left side equal to 0, and the solution will be the closed interval between these values.

#2x^2+9x+3<=0#
#2x^2+9x+3=0#

#x=frac{-9+-sqrt(9^2-4(2)(3))}{2(2)}#

#x=frac{-9+-sqrt57}{4}#

#x approx -0.362541#
#x approx -4.1374586#

Dec 22, 2016

Answer:

#x in (1/4(-sqrt57-9), 1/4(sqrt57-9))#
#= (-4.137, -0.3625)#, nearly. The segment of the x-axis in the shaded region of the graph illustrates the solution

Explanation:

Reorganizing,

#(x+9/4)^2+3/2-81/16<0#. So,

# |x+9/4| < sqrt57/4#. And so,

#1/4(-sqrt57-9)< x <1/4(sqrt57-9) #

graph{2x^2+9x+3 < 0 [-10, 10, -5, 5]}

Dec 22, 2016

Answer:

interval [-0.414, -0.36]

Explanation:

First solve this quadratic equation and find its 2 real roots by using the improved quadratic formula (Socratic Search);
#f(x) = 2x^2 + 9x + 3 <= 0#
#D = d^2 = b^2 - 4ac = 81 - 24 = 57# --> #d = +- sqrt57 = +- 7.55#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = - 9/4 +- 7.55/4#
#x1 = - 16.55/4 = - 4.14#
#x2 = -1.45/4 = - 0.36#.
Since a = 2 > 0, the parabola graph opens upward. Between the 2 real roots, f(x) < 0, as one part of the graph stays below the x-axis.
There for f(x) < 0 between the 2 real roots.
Answer by closed interval: [- 4.14, 0.36]
Both end points (-0.36 and -4.14) are included in the solution set.
Graph on the Number Line:

-------------------- -4.14 ============= -0.36 ---- 0 ----------------------