How to solve this inequality involving a 4th root?

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SUMMARY

The discussion focuses on solving the inequality involving the fourth root: \sqrt[4]{2x + 1} - 0.1 < \frac{1}{2}x + 1 < \sqrt[4]{2x + 1} + 0.1. The solution is established as -0.368935 < x < 0.677669. Participants suggest expanding \sqrt[4]{2x + 1} using binomial or Taylor series methods to simplify the inequality and recommend solving the inequalities separately for clarity.

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Homework Statement


[tex]\sqrt[4]{2x + 1} - 0.1 < \frac{1}{2}x + 1 < \sqrt[4]{2x + 1} + 0.1[/tex]


Homework Equations





The Attempt at a Solution



I'm having trouble getting just an 'x' by itself in the middle because of the 4th root. How should I solve this inequality? I tried everything but always end up with something to the 4th power after expanding 4 binomials.

(Solution: -0.368935 < x < 0.677669)
 
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PhizKid said:

Homework Statement


[tex]\sqrt[4]{2x + 1} - 0.1 < \frac{1}{2}x + 1 < \sqrt[4]{2x + 1} + 0.1[/tex]

I'm having trouble getting just an 'x' by itself in the middle because of the 4th root. How should I solve this inequality? I tried everything but always end up with something to the 4th power after expanding 4 binomials.

(Solution: -0.368935 < x < 0.677669)

I would think about expanding ##\sqrt[4]{1+2x}## as a binomial or Taylor series with error estimates.
 
Try to solve the inequalities separately.

[tex]\sqrt[4]{2x + 1} < \frac{1}{2}x + 1.1[/tex]
[tex]\sqrt[4]{2x + 1} > \frac{1}{2}x + 0.9[/tex]

ehild
 

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