How do you solve #m^4-2m^3-35m^2<=0# using a sign chart?

1 Answer
Sep 18, 2017

Answer:

#m^2(m^2-2m-35) <= 0# when #-5 <= m <= 7#

Explanation:

Let us factorize The LHS of #m^4-2m^3-35m^2 <= 0#

i.e. #m^2(m^2-2m-35) <= 0#

or #m^2(m-7)(m+5) <= 0#

Observe that as #m^2 >=0#, the above will satisfy only if

#(m-7)(m+5) < = 0#

Leaving equality for the time being,the product #(m-7)(m+5)# is desired to be negative.

It is apparent that sign of binomials #(m-7)# and #(m+5)# will change around the values #-5# and #7# respectively. In sign chart we divide the real number line using these values, i.e. below #-5#, between #-5# and #7# and above #7# and see how the sign of #(m-7)(m+5)# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)7#

#(m+5)color(white)(XXX)-ive color(white)(XXXX)+ive color(white)(XXXX)+ive#

#(m-7)color(white)(XXX)-ive color(white)(XXXX)-ive color(white)(XXXX)+ive#

#(m+5)(m-7)color(white)()+ive color(white)(XXX)-ive color(white)(XXXX)+ive#

It is observed that #(m+5)(m-7)<0# when #m# is between #-5# and #7#, which is the solution for the inequality. Wealso addequality sign as we have #m^2(m^2-2m-35) <= 0#

Hence #m^2(m^2-2m-35) <= 0# when #-5 <= m <= 7#