How do you solve the inequality #16x^2+8x+1>0# and write your answer in interval notation?

1 Answer
marfre
Mar 8, 2017

#(-oo, -1/4) U (-1/4, oo)#

Explanation:

Factor the parabola: #y = (4x + 1)(4x + 1) > 0#

Find out when the function #=0#: #(4x + 1) = 0#
#4x = -1# ; #x = -1/4#

Since both factors are the same: #y = (4x+1)^2#, #x = -1/4# is a touch point on the #x#-axis.

This means we have two choices for answers:
#(-oo, -1/4) U (-1/4, oo)# or no solution

To find out for sure, you can either graph the equation to see where the #y#-values are positive or you can select test points in the interval:
Let #x = -1#, #y = 9# so #(-oo, -1/4) > 0#

Let #x = 1#, #y = 25# so #(-1/4, oo) > 0#

Therefore the solution set in interval notation is: #(-oo, -1/4) U (-1/4, oo)#

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