How do you solve #x^4<4x^2#?

1 Answer
Jun 30, 2016

Answer:

#x in (-2, 0) uu (0, 2)#

i.e. #-2 < x < 0# or #0 < x < 2#

Explanation:

First note that if #x = 0# then #x^4 = 0 = 4x^2#, so the inequality is false.

If #x != 0# then #x^2 > 0# and we can divide both sides of the inequality by #x^2# to get:

#x^2 < 4#

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Case #bb(x > 0)#

Since #y = sqrt(x)# is a strictly monotonically increasing function for #x > 0#, we can take the square root of both sides of our simplified inequality to find:

#x < 2#

Hence we have solutions:

#0 < x < 2#

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Case #bb(x < 0)#

Note that #x^2 = (-x)^2# and #-x > 0#, so we find:

#-x < 2#

Multiplying both sides by #-1# and reversing the inequality, we get:

#x > -2#

Hence we have solutions:

#-2 < x < 0#

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Background

The truth or falsity of an inequality is unaltered by any of the following operations:

  • Add or subtract the same value from both sides.
  • Multiply or divide both sides by the same positive value.
  • Multiply or divide both sides by the same negative value and reverse the inequality (#<# becomes #>#, #>=# becomes #<=#, etc.).

More generally:

  • Apply the same strictly monotonically increasing function to both sides of the inequality.
  • Apply the same strictly monotonically decreasing function to both sides of the inequality and reverse the inequality.