Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I tried to figure out this question. Can any one help? I know the answer but I keep getting the wrong one. Thanks in advance.

Find all values of x that satisfy $|-(x + 1)^2+1|\geq 1$

share|improve this question
I have retagged your question. The tag (algebraic-geometry) is intended for questions in the field of Algebraic Geometry, a relatively advanced topic usually reserved for graduate courses. –  Alex Becker Jul 11 '12 at 3:12
For $M > 0, |x| \ge M \iff x \le -M$ or $x \ge M$ –  Vectk Jul 11 '12 at 3:19
First, you can remove the minus sign, as $|x|=|-x|$. –  Ross Millikan Jul 11 '12 at 3:20

6 Answers 6

$|x|\geq a\implies x\geq a$ or $x\leq -a$.Here, in your problem it results to, $1-(x+1)^2\geq 1\implies (x+1)^2\leq 0$, only solution for which is $x=-1.$ Also, there is a second case, $1-(x+1)^2\leq -1\implies (x+1)^2\geq 2\implies |x+1|\geq\sqrt 2\implies x\geq\sqrt 2-1$ or $x\leq -\sqrt2-1$.Therefore, the possible solutions of $x$ are $\{-1\}\cup(-\infty,-\sqrt2-1]\cup[\sqrt2-1,\infty)$ .

share|improve this answer
How did you get from (x+1)^2>=0 to x = -1. I thought it would be x >= -1 –  Mark Jul 11 '12 at 3:27
That is not $(x+1)^2\geq 0$ but $(x+1)^2\leq 0$ and since $(x+1)^2$ is a square, it can't be negative, hence $(x+1)^2=0\implies x=-1$. –  Aang Jul 11 '12 at 5:02

You can observe that $y=-(x+1)^2+1$ is a parabola with vertex in $(-1,1)$ and $a=-1$. Then you can draw its absolute value, and the line $y=1$.

If you solve the equation $(x+1)^2-1=1$ you will find the two intersection wich are, respectively, less than $-2$ and bigger than $0$ (i.e., $-1\pm\sqrt2$).

Finally you can write the solution: $x\le -1-\sqrt2 \vee x=-1 \vee x\ge-1+\sqrt2$.

enter image description here

share|improve this answer
$x=1$ is not the solution, $x=-1$ is. –  Aang Jul 11 '12 at 9:16
uh, right, I have fixed it, thank you –  zar Jul 11 '12 at 9:24

$|b-a|$ is the distance between $b$ and $a$ on the real line. So you look for $(x+1)^2 \leq 0 $ or $(x+1)^2\geq 2$.

share|improve this answer

$$|-(x + 1)^2+1|\geq 1 >0$$

Since both sides of the inequlity are positive squaring the inequality we get,

$$((x + 1)^2-1)^2 \geq 1 $$

$$\Leftrightarrow ((x + 1)^2-1)^2-1^2\geq 0 \tag {difference of squares} $$

$$\Leftrightarrow ((x + 1)^2-1-1)((x + 1)^2-1+1)\geq 0 $$

$$\Leftrightarrow ((x + 1)^2-2)(x+1)^2 \geq 0 \tag {with $(x+1)^2>0$ iff $x \neq -1$} $$


Case 1 $x = -1$

Then, $((x + 1)^2-2)(x+1)^2 =0 \geq 0$ as required

Case 2 $x \neq -1$

Then $(x + 1)^2-2\geq 0$ (dividing the inequality by $(x + 1)^2>0$)

$$\Leftrightarrow(x + 1- \sqrt{2})(x + 1+ \sqrt{2})\geq 0$$

$$ \Leftrightarrow x \leq -1-\sqrt{2} \ or \ x \geq -1+\sqrt{2}$$

Therefore the final solution is $$x \in (- \infty,-1-\sqrt{2} ] \cup \{-1 \} \cup [-1+\sqrt{2}, +\infty)$$

share|improve this answer

A very slightly different approach to the ones already given is to recall that $|y| = \sqrt{y^2}$. So we need to find $x \in \mathbb{R}$ such that

$$ \sqrt{\left(1-(x+1)^2\right)^2} \geq 1 $$

Squaring both sides you get

$$ \left(1-(x+1)^2\right)^2 \geq 1 \quad \Longrightarrow \quad -2(x+1)^2 + (x+1)^4 \geq 0 $$

If $x \neq -1$ (check that it is also a solution), we can divide both sides by $(x+1)^2$ and fall onto a quadratic equation that is easy to solve. Finally, the solution set is $$ \left\lbrace x \in \mathbb{R} \;|\; x \leq -1-\sqrt{2} \;\;\text{or}\;\; x=-1 \;\;\text{or}\;\; x \geq -1+\sqrt{2} \right\rbrace $$

share|improve this answer

Observe that $(x+1)^2\geq 0$, so $1-(x+1)^2\leq 1$, with equality if and only if $(x+1)^2=0$ if and only if $x+1=0$ if and only if $x=-1$. Thus, $x=-1$ is a solution to the inequality.

The only other option to satisfy it is when $1-(x+1)^2\leq -1$ if and only if $-x^2-2x\leq -1$ if and only if $0\leq x^2+2x-1$. Starting with equality, we find from the quadratic equation that $0=x^2+2x-1$ if and only if $x=-1\pm\sqrt{2}$. Using test points in the intervals $(-\infty,-1-\sqrt{2})$, $(-1-\sqrt{2},-1+\sqrt{2})$, $(-1+\sqrt{2},\infty)$, it is readily seen that $x^2+2x-1$ is positive on the first and last interval and negative on the middle. Thus, $x^2+2x-1\geq 0$ if and only if $x\leq -1-\sqrt{2}$ or $x\geq -1+\sqrt{2}$.

Consequently, the solution set is $\left(-\infty,-1-\sqrt{2}\right]\cup\{-1\}\cup\left[-1+\sqrt{2},\infty\right)$.

share|improve this answer
the points $-1-\sqrt 2$ and $-1+\sqrt 2$ should be included in solution set as they satisfy the given inequality. –  Aang Jul 11 '12 at 5:01
D'oh! Closed rays. Thanks. –  Cameron Buie Jul 11 '12 at 13:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.