How do you solve the inequality #x^2<=abs(4x-3)# and write your answer in interval notation?

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Answer 1 · 2017-10-01T17:34:22

#x^2 <= abs(4x - 3)#

so, we have either #x^2 <= 4x - 3# or #x^2 <= -(4x - 3)#

i.e. #x^2 - 4x + 3 <= 0# or #x^2 + 4x - 3 <= 0#

solving #x^2 - 4x + 3 = 0#, we #x = 1 or x = 3#

Hence, #1 <= x <= 3#

Solving #x^2 + 4x - 3 = 0# using the quadratic formula gives:

#x = -2 +- sqrt7#

Hence, #-2 - sqrt7 <= x <= -2 + sqrt7#

Using interval notation we have:

#[-2 - sqrt7, -2 + sqrt7]# OR #[1, 3]#

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