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

Question

1654 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-15T09:45:10

We first consider where #x^2 - 4x = 0#

i.e. when #x(x - 4) = 0#

Hence, #x = 0 or x = 4#

We then need to consider three seperate regions.

i.e. #x < 0#, #0 < x < 4# and #x > 4#

For #x < 0#, we have that #x^2 - 4x > 0#

For #0 < x < 4#, we have that #x^2 - 4x < 0#

and for #x > 4# we have that #x^2 - 4x > 0#

Hence, for #x^2 - 4x >=0# we have two regions.

i.e. #x <=0# or #x >=4#

:)>