The problem asks us to find the solutions for the following equations: a) $(0-3)x^2 = 0$ b) $(3-\sqrt{2})x^2 = 0$ c) $mx^2 = 0$ for $m \neq 0$ d) $(m-1)x^2 = 0$ for $m=1$

1. Problem Description

The problem asks us to find the solutions for the following equations:

a) $(0-3)x^2 = 0$

b) $(3-\sqrt{2})x^2 = 0$

c) $mx^2 = 0$ for $m \neq 0$

d) $(m-1)x^2 = 0$ for $m=1$

2. Solution Steps

a) $(0-3)x^2 = 0$

$-3x^2 = 0$

$x^2 = \frac{0}{-3}$

$x^2 = 0$

$x = 0$

b) $(3-\sqrt{2})x^2 = 0$

Since $3 - \sqrt{2} \neq 0$, we can divide both sides by $(3-\sqrt{2})$:

$x^2 = \frac{0}{3-\sqrt{2}}$

$x^2 = 0$

$x = 0$

c) $mx^2 = 0$ for $m \neq 0$

Since $m \neq 0$, we can divide both sides by $m$:

$x^2 = \frac{0}{m}$

$x^2 = 0$

$x = 0$

d) $(m-1)x^2 = 0$ for $m=1$

Substitute $m=1$ into the equation:

$(1-1)x^2 = 0$

$0x^2 = 0$

$0 = 0$

This equation is true for any value of $x$. Thus, $x$ can be any real number.

3. Final Answer

a) $x = 0$

b) $x = 0$

c) $x = 0$

d) $x \in \mathbb{R}$