Error analysis: You are going to find the error(s) in the math work shown below. Explain any errors completely, then enter the correct solution including the correct steps. For both problems thank you

[tex]\sf \sqrt{\dfrac{15}{2} } = \dfrac{\sqrt{15} }{\sqrt{2} }[/tex]
[tex]\sf = \dfrac{\sqrt{15} }{\sqrt{2} } \ \cdot \ \dfrac{\sqrt{2} }{\sqrt{2} }[/tex]
[tex]\sf = \dfrac{\sqrt{15 * 2} }{\sqrt{2} * \sqrt{2} }[/tex] [ apply radical rule: √a × √b = √ab ]
[tex]\sf = \dfrac{\sqrt{30} }{2}[/tex] [multiply the integers] [note: √2 × √2 is 2]
[tex]\sf x^2 -4 = 0 \\\\\sqrt{x^2-4} = 0\\ \\x^2 - 2^2 = 0^2 \ \quad \quad \quad \leftarrow third \ step \ should \ be \ this\\\\ (x-2)(x + 2) = 0 \\\\x = 2, x = -2 \\ \\ x = \pm 2[/tex]
Answer:
2. (√30)/2
3. x = ±2
Step-by-step explanation:
For these, we will show the correct steps, and describe the error in words.
__
[tex]\displaystyle\begin{aligned}\sqrt{\dfrac{15}{2}&=\dfrac{\sqrt{15}}{\sqrt{2}}\\&=\dfrac{\sqrt{15}}{\sqrt{2}}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}}\\ &=\dfrac{\sqrt{30}}{2}&\text{error in original work was here}\end{aligned}[/tex]
The product of √2 and itself is 2, not √2.
__
[tex]\displaystyle\begin{aligned}x^2-4&=0\\x^2&=4&\text{error was here}\\x&=\pm\sqrt{4}&\text{both square roots}\\x&=\pm2\end{aligned}[/tex]
The expression (x²-4) does not have a square root of (x-2). To usefully take the square root, we need a perfect square involving the variable term.