How do you solve #(24-w)/w <= 3w-8#?

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Answer 1 · 2017-05-13T04:32:59

#w>=(7+sqrt337)/6# or #(7-sqrt337)/6<=w<0# and in interval notation #[(7-sqrt337)/6,0)uu[(7+sqrt337)/6,oo)#

#(24-w)/w<=3w-8#

It is apparent that we cannot have #w=0#

Assuming #w>0# we have #24-w<=w(3w-8)#

or #24-w<=3w^2-8w#

or #3w^2-7w-24>=0#

Quadratic formula gives solution of equality as #w=(7+-sqrt(49+288))/6# i.e. #w=(7+-sqrt337)/6#

and hence for inequality we have #w>=(7+sqrt337)/6#

Note that we already have a condition that #w>0#

In case #w<0#, we have #24-w>=w(3w-8)#

i.e. #3w^2-7w-24<=0#

i.e. #(7-sqrt337)/6<=w<0#

and in interval notation #[(7-sqrt337)/6,0)uu[(7+sqrt337)/6,oo)#