How do you solve the inequality #5x+4<=3x^2# and write your answer in interval notation?

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Answer 1 · 2017-12-11T10:52:02

The solution is # x in (-oo, -0.59] uu[2.26, +oo)#

Let's rewrite the inequality

#5x+4<=3x^2#

#3x^2-5x-4>=0#

Let #f(x)=3x^2-5x-4#

The roots of the quadratic equation #3x^2-5x-4=0#, are

# x = ( 5+-sqrt ((-5)^2-(4)* (3) * (-4)) ) /(6)=(5+-sqrt(73))/(6) #

#x_1=(5-sqrt73)/6=-0.59#

#x_2=(5+sqrt73)/6=2.26#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaaaaa)##x_1##color(white)(aaaaaa)##x_2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-x_1##color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-x_2##color(white)(aaaaa)##-##color(white)(aaaa)####color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##0##color(white)(aaa)##-##color(white)(aa)##0##color(white)(aa)##+#

Therefore,

#f(x)>=0# when # x in (-oo, x_1] uu[x_2, +oo)#

graph{3x^2-5x-4 [-11.39, 11.11, -6.615, 4.635]}