How do you solve and write the following in interval notation: #4x^2 + x ≤ 3#?

1 Answer
Jun 10, 2016

Answer:

Solution is #-1<=x<=3/4#.

Explanation:

#4x^2+x<=3# can be written as #4x^2+x-3<=0#

Now factorizing LHS, we get

#4x^2+4x-3x-3<=0#

or #4x(x+1)-3(x+1)<=0#

or #(4x-3)(x+1)<=0#

As equality sign is fulfilled by #x=-3/4# and #x=-1# and these are part of solution as we already have equality sign.

These two points divides number line in three regions

A- First region is #x<-1# - In this region both #(4x-3)# and #(x+1)# are negative and as such #4x^2+x-3>0#, hence this is not a solution.

B- Second region is #-1<x<3/4# - In this region while #(x+1)# is positive, #(4x-3)# is negative and as such #4x^2+x-3<0#, hence this is a solution.

C- Third region is #3/4<x# - In this region while #(x+1)# and #(4x-3)# both are positive and as such #4x^2+x-3>0#, hence this is a not a solution.

Hence solution is #-1<=x<=3/4#.