in $x^2=9$, how exactly do we conceive of taking the square root function when translating it to $x=\pm\sqrt(9)$? do we imagine the ordinary square root function applied, then the $\pm$ symbol applied, or do we conceive of a special function $\pm \sqrt{}$, which in one function call obtains both the positive and negative root? My main question is how we conceive of it for inequalities. For the inequality $x^2<9$, when we take the square root of both sides,when we consider the negative root we flip the inequality sign- so there are two cases $x<3$, $x>-3$. How do we conceive of what the square root function does when we apply it to both sides? How can we conceive of how this is related to the flipping of the inequality sign? if we write this as $x<\pm 3$, this would be wrong as $x>-3$ with the inequality sign flipped for the negative case- so applying the square root to both sides is different for inequalities as it is for equations- why is this case; why can't we just take the square root of both sides like we did for the equation in the equality?
thanks
How taking plus or minus square root works, specifically for inequality
inequality
Best Answer
We start from $$x^2=9.$$ Then apply square root to both sides $$\sqrt{x^2}=\sqrt{9}$$ $$\sqrt{x^2}=3.$$ But $$\sqrt{x^2}=|x|.$$ So $$|x|= 3$$ $$x = \pm 3.$$
The nontrivial part here is $\sqrt{x^2} = |x|$. How do we know that these two functions are equal? Well, we can check that they are equal for two separate cases: $x\geq 0$ and $x<0$.
For the inequality it is similar. $$x^2 <9$$ $$\sqrt{x^2}<\sqrt{9}$$ (we can do above, because both sides are positive, so applying square root function keeps the inequality(because square root function is increasing)
$$|x|<3.$$ And how do we solve this? Well, equations/inequalities with absolute value tend to be solved by cases. If $x\geq 0 $, then $|x|=x$ and so $$x<3 \text{ and } x\geq 0$$ which gives first set of solutions $$0 \leq x < 3 .$$ If $x<0$, then $|x|=-x$ and so $$-x<3 \text{ and } x< 0$$ $$x>-3 \text{ and } x< 0$$ which gives second set of solutions $$-3 < x < 0 .$$
Connecting these two solutions $0\leq x < 3$ and $-3 < x < 0$ we get the final answer $-3< x < 3$.