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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following inequality: $$|4 - k^2| > |10 + 13k|$$ So how to solve this ?

share|cite|improve this question
Is $k$ real or complex? – Daniel Fischer Jun 8 '14 at 14:58
@DanielFischer $k$ is real – hbak Jun 8 '14 at 14:59
Then a case distinction would be an easy way to find all possible $k$. – Daniel Fischer Jun 8 '14 at 15:00
@DanielFischer , I know that my question is so simple, but i stick with this problem , could you help me ? – hbak Jun 8 '14 at 15:01
Consider the cases $k \geqslant 2$, $-\frac{10}{13} \leqslant k < 2$, $-2\leqslant k < -\frac{10}{13}$ and $k < -2$ separately. – Daniel Fischer Jun 8 '14 at 15:03


Start by sketching the graphs $y=|4-x^2|$ and $y=|10+13x|$. To do this, sketch $y=4-x^2$ and $y=10+13x$ and reflect anything below the $x$-axis back up above the $x$-axis.

Next, solve the two equations $4-x^2 = \pm(10+13x)$.

Use your sketch to help you find the regions where $|4-x^2| > |10+13x|$.

share|cite|improve this answer

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.