How do you solve #3x^2-16x+5<=0# by algebraically?

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Answer 1 · 2018-05-12T06:02:33

The solution is # x in [1/3, 5]#

First factorise the equation

#3x^2-16x+5=(3x-1)(x-5)<=0#

#(3x-1)(x-5)=0#

when

#{(3x-1=0),(x-5=0):}#

#{(x=1/3),(x=5):}#

Let #f(x)=(3x-1)(x-5)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaaa)##1/3##color(white)(aaaaaaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##3x-1##color(white)(aaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##x-5##color(white)(aaaaa)##-##color(white)(aaaa)####color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

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

Therefore,

#f(x)<=0# when # x in [1/3, 5]#