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When answer this kind of inequality $|2x^2-5x+2| < |x+1|$ I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -. When I check the negative options, I need to flip the inequality sign? Thanks

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  • $\begingroup$ Let's see... roots are $$x=\frac{5\pm\sqrt{25-16}}{4}=\frac{5\pm 3}{4}=2\text{ or }1/2.$$ $\endgroup$ – Brady Trainor Oct 28 '13 at 10:43
  • $\begingroup$ Or another strategy is check when they're equal, which is only two cases (of the $\pm$ bit that is). But your strategy would probably work, I was just fishing for a trick. And you only flip the inequality when you multiply both sides by negative one. Be careful with your cases. $\endgroup$ – Brady Trainor Oct 28 '13 at 10:48
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Another way - no tricks, just systematically looking at all cases, where we can write the inequality without the absolute value sign, to convince you that all possibilities are covered..

Following the definition of the absolute value function, RHS is easy to rewrite as, $$|x+1| = \begin{cases} x+1 & x \ge -1\\ -x-1 &x < -1 \end{cases}$$

Noting $2x^2-5x+2 = (2x-1)(x-2)$, we have the LHS as $$|2x^2-5x+2| = \begin{cases} 2x^2-5x+2 & x \le \frac12 \text{ or } x \ge 2\\ -2x^2+5x-2 &\frac12 < x < 2 \end{cases}$$

Thus we can break the number line into for regions (using break points at $x = -1, \frac12, 2$) and write the inequality in each region without any absolute value symbols. Then we solve the inequality in each region, and combine results in the end. Details follow.

Region 1: $x < -1$
Here the inequality is $2x^2-5x+2 < -x - 1$ which is equivalent to
$2x^2 - 4x + 3 < 0$ or $2(x-1)^2+1 < 0$, which is not possible.

Region 2: $-1 \le x \le \frac12$
Here we have the inequality as $2x^2-5x+2 < x + 1 \iff 2x^2 - 6x + 1 < 0$ which is true when $\frac12(3-\sqrt7) < x < \frac12(3+\sqrt7)$. Taking the part of this solution which is in the region being considered, we have solutions for $x \in (\frac12(3-\sqrt7), \frac12]$

Region 3: $\frac12 < x < 2$
Here the inequality is $-2x^2+5x-2 < x + 1 \iff 2(x- 1)^2 + 2 > 0$ which holds true for all $x$. So the entire region is a solution.

Region 4: $2 \le x$
Here the inequality is $2x^2-5x+2 < x + 1$, which is exactly the same form as Region 2, and has the same solutions. Hence $2 \le x < \frac12(3+\sqrt7)$ is the solution from this Region.

Combining solutions from all regions, we have $\frac12(3-\sqrt7) < x < \frac12(3+\sqrt7)$ or $|x - \frac32| < \frac{\sqrt7}{2}$ as the complete solution set (which others have already pointed out).

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  • $\begingroup$ What happens with the points at the boundary? Why did you chose $x\le -1$ and $x>-1$ and not $x<-1$ and $x\ge-1$. Did you check the values by plugging in the boundary values? $\endgroup$ – Alexander Cska Feb 18 '18 at 17:57
  • $\begingroup$ @AlexanderCska Check the answer carefully , all boundary cases are also covered in the regions defined. You should work out yourself what would happen if the regions were chosen differently, it is not difficult, and in this case does not matter at all. $\endgroup$ – Macavity Feb 18 '18 at 18:00
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An inequality of the form $|A|<B$ is equivalent to the pair of inequalities $-B < A < B$. (If $B$ is negative, then they are still equivalent - it's just easy to see that neither has any solutions.) So your inequality $|2x^2−5x+2|<|x+1|$ is equivalent to $$ -|x+1|<2x^2−5x+2<|x+1|. $$ Now $|B|=B$ if $B$ is positive, while $|B|=-B$ if $B$ is negative. Therefore, when $x+1>0$ (equivalently, when $x>-1$), this becomes $$ -(x+1)<2x^2−5x+2<x+1 \quad\text{(when $x>-1$),} $$ while the second case is $$ x+1<2x^2−5x+2<-(x+1) \quad\text{(when $x<-1$)}. $$ It's not hard to see that $-(x+1)<2x^2−5x+2$ always, so the second case never occurs. For the first case, it remains to decide when both $2x^2−5x+2<x+1$ and $x>-1$ occur simultaneously; the answer turns out to be $x\in \left(\frac{1}{2} \left(3-\sqrt{7}\right),\frac{1}{2} \left(3+\sqrt{7}\right)\right)$.

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  • $\begingroup$ How did you get to the second case? And why it's never occurs? $\endgroup$ – gbox Oct 28 '13 at 12:27
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if you take $x<0$

$-(2x^2-5x+2)<-(x+1)$

$-2x^2+6x-1<0$

you gets roots:

$x_1=\frac {1}{2}(3−7\sqrt7)≈2.82$

$x_2=\frac {1}{2}(3+\sqrt7)≈0.18$

Becouse we have x<0 (x is negative and our roots are pozitive) there are solutions:

$x \in { \emptyset}$

if $X>0$

$2x^2 - 5x + 2 < x +1$

$2x^2 - 6x + 1 < 0$

you gets roots:

$ x_1 = \frac {1}{2}( 3-\sqrt 7) \approx 0.18 $

$ x_2 = \frac {1}{2}( 3+\sqrt7) \approx 2.82$

so you get

$ \frac {1}{2}( 3-\sqrt7) \leq x \leq \frac {1}{2}( 3+\sqrt7) $

if we combine both results we get

$x \in {( \frac {1}{2}( 3-\sqrt 7), \frac {1}{2}( 3+\sqrt 7) )}$

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  • $\begingroup$ but what about $x>0$? and why did you made x negtaive and not all the expression $-(x+1)$ $\endgroup$ – gbox Oct 28 '13 at 10:48
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Let me throw in my 5 cent. This type of inequality is of a very special type, the solution of which follows from the fact that you are allowed to square both sides, since the absolute value can't be negative. Therefore $$ \left(\left|2x^2-5x+2\right|\right)^2<\left(\left|x+1\right|\right)^2 $$ Here the absolute values can be simply dropped. Doing so and applying the following formula $$ a^2-b^2=(a-b)(a+b) $$ yields $$ \left( 2x^2-6x+1 \right)\left(2x^2-4x+3 \right)<0. $$

The second term $2x^2-4x+3 $ represents a parabola, which never crosses the x-axis, since the discriminant is negative. Therefore all you need is to check for which values of $x$ the first term gets negative. Factoring the first quadratic equation yields
$$ \left( x- \frac{3-\sqrt{7}}{2} \right)\left(x- \frac{3+\sqrt{7}}{2}\right)\left(\text{Term} >0: \forall x \in \mathbb{R}\right)<0 $$

Therefore the solution is $$ x\in\left(\frac{3-\sqrt{7}}{2},\frac{3+\sqrt{7}}{2}\right) $$

P.S. The formula $$ g(x)<|f(x)|<-g(x) $$

works quick and fine mainly in cases where a single absolute value is present one side of the inequality. Say if we had $$ \left|2x^2-5x+2\right|<x+1 $$

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