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

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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

1 Answer 1


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 $|4-x^2| > |10+13x|$.

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