Solve the following equation: \( b^{2}=3 b \) \( \begin{array}{l}\{3,-3\} \\ \text { \{3, 0\} } \\ \text { \{-3, 0\} } \\ \text { \{3\} }\end{array} \)

To solve the equation \( b^{2} = 3b \), we can rearrange it to \( b^{2} - 3b = 0 \). Factoring out a \( b \), we get \( b(b - 3) = 0 \). This gives us two solutions: \( b = 0 \) and \( b = 3 \). Therefore, the solution set is \( \{3, 0\} \). Here’s a fun tip: when you see an equation like this, think of it as a sliding puzzle where you’re trying to get \( b \) to one side equal to the other! The zero product property comes in handy here, since if one factor is zero, the entire product equals zero. And just a heads up! A common mistake when factoring is forgetting to set each factor to zero. Always remember: if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)!