Problem with a simple exercise So I have to solve the equation $$y^2=4\tag{1.9.88 unit 3*}$$
I did this: $$y^2=4 \text{ means } \sqrt{y^2}=\sqrt{4}=>y=2$$
But I have a problem, $y$ can be either negative or positive so I need to do: $$\sqrt{y^2}=|y|=2=>y=2- or- y=-2$$
Is it right?
 A: Yes, it is right. I'll recommend a better way to approach this. Just factorize it.
$(y-2)(y+2) = 0$
$y = 2,-2$
A: Never forget this rule: For all $x\in\mathbb{R}$, the following holds: $$\sqrt{x^2}=|x|=\begin{cases}x & x\gt 0\\ -x & x\leqslant0\end{cases}$$
Applying that to the equation $y^2=4:$ $$\begin{align}
\sqrt{y^2}=\sqrt{4}&\iff |y|=2\\ &\iff y=2\,\,\mathrm{or}\,\,y=-2
\end{align}$$
So you're right, keep up the good work!
I hope this helps.
Best wishes, $\mathcal H$akim.
A: For this type of problem, I always think, "How would it appear on a graph? What are the x-intercepts?" Of course, I do not have time to actually draw it out. But I can think, "Let $y^2$ be $x^2$. How can $x^2=4$ be turned into a function? I can just move $4$ to the left by subtracting $4$ on both sides, and replacing the $0$ with $f(x)$."
So, what are the x-intercepts of $f(x)=x^2-4$? Hopefully you can immediately identify the x-intercepts as $\pm \ 2$. So that means $x=\pm \ 2$ if $x^2=4$. Just replace $x^2$ with $y^2$. If $y^2=4$, then $y=\pm \ 2$.
Another much quicker way to solve the problem is factoring. Subtract $4$ from both sides to get $y^2-4 = 0$. Then factor the equation using difference of squares. $(y+2)(y-2)=0$. The equation now splits into two different cases. You know that if $ab=0$, then either $a=0$, $b=0$, or $a$ and $b$ both equal $0$.
Case 1: $y+2=0$
$y+2=0$
$y=-2$
Case 2: $y-2=0$
$y-2=0$
$y=2$
So, the answer is $y = \pm \ 2$.
Extra Information Below:
You may be wondering, "Why can you just replace the $0$ in $x^2-4=0$ with $f(x)$ in the first method of solving the equation?" You must know the difference between $f(x)=x^2-4$ and $0=x^2-4$. $f(x)=x^2-4$ is the graph of all the points that satisfy the condition that each point on the curve has coordinates $(x, f(x))$. $x^2-4=0$ is a special case of the function $f(x)=x^2-4$. Here, $0=f(x)$. So, what points on the curve are in the form $(x, 0)$? Obviously, the points are $(2, 0)$ and $(-2, 0)$.
You can also think of the equation $x^2-4=0$ as a system of equations like:
$$\begin{cases} f(x)=x^2-4 \\ f(x)=0 \\ \end{cases}$$
The solution to the system will be the intersections of the function $f(x)=x^2-4$ and the line $f(x)=0$. The line $f(x)=0$ is the x-axis, so the question becomes, "Where does $f(x)=x^2-4$ intersect with the x-axis?" Obviously, they intersect at the points $(2, 0)$ and $(-2, 0)$.
