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We all know that if we want to solve absolute value inequality with two absolute values on both sides we have to squaring them

Like this ->>>

$\begin{aligned} |2x+1| &\ge|x-2| \\ (2x+1)^2 &\ge (x-2)^2 \\ \end{aligned}$

But how if i want to solve them "without squaring both sides" I have read some books and theory about solve them without squaring but i still don't understand. If you know some, please feel free to tell me. Thanks

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First, set

$$ 2x+1 = \pm (x-2)$$

to obtain the ‘break’ points. They are $x=-3$ and $x=1/3$. This is an important step for dealing with multiple absolute terms appearing in the inequality.

Then, check whether, in each of the three regions separated by the break points, the inequality holds. This could be easily done with some convenient check points. For example, $-5$ for the left region, $0$ for the middle, and $1$ for the right, as shown below,

$$ |2(-5)+1|\ge |-5-2|, \space \text{true}$$

$$ |2\times 0+1|\ge |0-2|, \space \text{false}$$

$$ |2\times 1+1|\ge |1-2|, \space \text{true}$$

The solutions then follow

$$ x \le -3, \space \space x \ge 1/3$$

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Do you know for example how to solve $|2x+1| \ge 5$ without squaring both sides?

This example has the solution $2x+1 \ge 5$ or $2x+1 \le -5$, (If you want to understand this just take some numbers in these 2 intervals and check if there absolute value will be greater than $5$ or not, and yes it will be greater than $5$)

And to solve for example $|x-2| \le 3$, the solution is $-3\le x-2 \le 3$

Do you know now how to solve your inequality?

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  • $\begingroup$ I do. Thanks. So it's use the basic property of absolute value right. $\endgroup$ – user516076 Aug 23 '19 at 3:50
  • $\begingroup$ Yes that is true $\endgroup$ – Fareed Abi Farraj Aug 23 '19 at 3:54

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