How do you solve #x(3-x)(x-5)<=0#?

1 Answer
Shwetank Mauria
Jul 31, 2016

Solution is #0<=x<=3# or #x<=5#.

Explanation:

As #x(3-x)(x-5)<=0#, we have following options

either

(1) #x<0# - in such a case #x# and #x-5# are negative and #3-x# is positive and hence #x(3-x)(x-5)# is positive and hence #x<0# is not a solution.

(2) #0<=x<=3# - in such a case #x# and #3-x# are positive and #x-5# is negative and hence #x(3-x)(x-5)# is negative and hence #0<=x<=3# is a solution .

(3) #3< x< 5# - in such a case #x-5# and #3-x# are negative and #x# is positive and hence #x(3-x)(x-5)# is positive and hence #3< x< 5# is not a solution .

(4) #x<=5# - in such a case #x# and #x-5# are positive and #3-x# is negative and hence #x(3-x)(x-5)# is negative and hence #0<=x<=3# is a solution.

Hence solution is #0<=x<=3# or #x<=5#.

graph{x(3-x)(x-5) [-10, 10, -10, 10]}

Related questions
See all questions in Polynomial Inequalities
Impact of this question
2048 views around the world
You can reuse this answer
Creative Commons License