Question

what are the solutions of $12 - x^2 = 0$?
\\(\bigcirc\\) $x = 2\sqrt{3}$ and $x = -2\sqrt{3}$
\\(\bigcirc\\) $x = 3\sqrt{2}$ and $x = -3\sqrt{2}$
\\(\bigcirc\\) $x = 4\sqrt{3}$ and $x = -4\sqrt{3}$
\\(\bigcirc\\) $x = 6$ and $x = -6$

Explanation:

Step1: Rearrange the equation

We start with the equation \(12 - x^{2}=0\). Let's rearrange it to isolate \(x^{2}\). Add \(x^{2}\) to both sides and subtract 0 from both sides (which doesn't change the equation) to get \(x^{2}=12\).

Step2: Solve for \(x\)

To solve for \(x\), we take the square root of both sides. Remember that when we take the square root of a number, we get both a positive and a negative solution. So, \(x = \pm\sqrt{12}\). Now, simplify \(\sqrt{12}\). We can factor 12 as \(4\times3\), and since \(\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}\) (by the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) for \(a\geq0,b\geq0\)), and \(\sqrt{4} = 2\), we have \(\sqrt{12}=2\sqrt{3}\). So, \(x=\pm2\sqrt{3}\), which means \(x = 2\sqrt{3}\) and \(x=-2\sqrt{3}\).

Answer:

\(x = 2\sqrt{3}\) and \(x = -2\sqrt{3}\) (the first option)