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I have a question about why the unknown becomes absolute when taking the square root in an inequality. For example:

Find the value(s) of $k$ for which the equation $2x^2-kx+3=0$ will have two different real roots.

$$b^2-4ac>0$$ $$(-k)^2-4(2)(3)>0$$ $$k^2-24>0$$ $$k^2>24$$ The step afterwards is the step I do not understand. $$|k|>\sqrt{24}$$ Why does $k$ become an absolute value? Also, why isn't there a $\pm$ sign at the front of $\sqrt{24}$?

The steps are continued below:

$$|k|>2\sqrt6$$ $$-2\sqrt 6>k>2\sqrt 6$$

I have had a look at a similar question asked before, "Taking the square roots in inequalities", but I don't really understand it.

Also, why does $|k|>2\sqrt6 $ become $-2\sqrt 6>k>2\sqrt 6$?

Thank you.

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The given final step is wrong as you have identified.

It should be as follows:

$$|k|>2\sqrt{6} \Rightarrow k>2\sqrt{6} \,\, \text{or,} \,\, k<-2\sqrt{6}$$

And as far as your question about absolute values goes, $$k^2>24$$ is valid for both positive and negative values of $k$ within a certain range and the $\pm$ sign cannot be used in case of inequalities in the way they are used in equations, since the treatment is different for the $2$ different signs.

For example, $k>\pm 24$ does not imply $k>2\sqrt{6} \,\, \text{or,} \,\, k<-2\sqrt{6}$ which is the answer in this case.

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  • $\begingroup$ Thanks. Just one question about the final step, what is the difference between $k>2\sqrt 6$ or $k<-2\sqrt 6$ and $-2\sqrt6>k>2\sqrt6$? $\endgroup$ – Squidiemeister Dec 27 '15 at 1:07
  • $\begingroup$ The first inequality is correct while the second is wrong. Since in the second one, it implies $-2\sqrt{6}>2\sqrt{6}$ which I believe you know what it means.. $\endgroup$ – SchrodingersCat Dec 27 '15 at 6:45
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The absolute values take care of the $\pm$ on the square root. You're correct that when you take a square root, you would expect $\pm\sqrt{24}$. The problem here is that the inequality gets in the way. If you ignore the inequality, you're left with $k>\pm\sqrt{24}$, which isn't what you intend. The problem is that, depending on if $k$ is positive or negative, the sign of the inequality changes.

More precisely, if $k^2>24$ and $k$ is positive, then it must be that $k>\sqrt{24}$ (if $0<k\leq\sqrt{24}$, then the square of $k$ will be at most $24$).

On the other hand, if $k^2>24$ and $k$ is negative, then it must be that $k<-\sqrt{24}$. The proof of this is not very enlightening, but if we cheat and play with the statement $k<-\sqrt{24}$ and multiply both sides by $k$, since $k$ is negative, the inequality reverses so that $k^2>-k\sqrt{24}$ and that multiplying both sides of $k<-\sqrt{24}$ by $-\sqrt{24}$ gives $-k\sqrt{24}>24$. Combining these inequalities gives $k^2>24$ (even though this isn't a proof).

The two inequalities $k>\sqrt{24}$ or $-k>\sqrt{24}$ combine to show that $|k|>\sqrt{24}$ because $|k|=k$ when $k$ is positive and $|k|=-k$ when $k$ is negative.

You can think about it this way, since $k^2=|k|^2$, the sign of $k$ doesn't matter when you square $k$, only its magnitude. Therefore, for $k^2>24$, it must be that $k$ is either a big positive or a big negative number (the sign disappears under squaring, but the magnitude must be large enough so that the square is at least $24$).

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