[Math] Using the absolute value when taking the square root in an inequality

algebra-precalculusquadratics

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.

Best Answer

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.

Related Question