How do you solve #4x^2-8x>=0# using a sign chart?

1 Answer
Jan 25, 2017

Answer:

The answer is #x in ]-oo, 0] uu [2, +oo [#

Explanation:

Let s factorise the expression

#4x^2-8x=4x(x-2)#

Let #f(x)=4x(x-2)#

The domain of #f(x)# is #D_f(x)=RR#

Now we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##0##color(white)(aaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

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

Therefore,

#f(x)>=0# when #x in ]-oo, 0] uu [2, +oo [#