For some reason, I'm having a hard time solving for $x$ in this equation: $x^2=y,-2\lt x \lt 3.$ I could use some help. Thanks.
Solving for $x$ in terms of $y$
2 Answers
Without the constraint $-2 < x <3$, the solution to $x^2 = y$ is given by $x = \sqrt{y} \text{ or } - \sqrt{y}$ whenever $y \geq 0$. The constraint $-2 < x <3$ demands that $\sqrt{y} \in [0,3)$ and $-\sqrt{y} \in (-2,0]$. Hence, this translates into $y \in [0,9)$ and $y \in [0,4)$ respectively.
Hence, we now need to split this into cases.
First we need $y \geq 0$. Else there are no solutions. Hence, we assume $y \geq 0$.
Next, if $y < 4$, then $x = \sqrt{y}$ or $x = -\sqrt{y}$. Note that since $y <4$, we have that $\pm \sqrt{y} \in (-2,3)$.
Next, if $4 \leq y < 9$, then $x = \sqrt{y}$. This is so since $-\sqrt{y} < -2$ whenever $4 \leq y < 9$. But $\sqrt{y} \in [2,3) \subset (-2,3)$, since $4 \leq y < 9$.
If $y \geq 9$, then $\sqrt{y} \geq 3$ and $-\sqrt{y} \leq -3$. Hence, no solution.
Hence, to summarize $x = \begin{cases} \text{No solution} & \text{ if }y<0\\ \sqrt{y} \text{ or } -\sqrt{y}& \text{ if }0 \leq y < 4\\ \sqrt{y} & \text{ if }4 \leq y < 9\\ \text{No solution} & \text{ if } y \geq 9 \end{cases}$
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0Thanks. the motivation certainly helped! :) – 2012-06-04
First thing one notices in such cases is that when $y=x^2$ means $y$ cannot be negative. So we'll keep in mind that $y \geq 0$.
Since $-2
But what will happen When $ 0 \leq y <4$ or $y \geq 9$?
$ 0 \leq y <4 \implies 0 \leq x^2 <4 \implies -2
$y \geq 9\implies x^2 \geq 9 \implies x \geq 3$ and since it has no intersection with $-3
Finally,
$x=\pm \sqrt y, 0 \leq y <9 $