- Identify the common factor \(6w\) in the expression \(6w^2 - 30w\).
- Factor the expression to get \(6w(w - 5) = 0\).
- Apply the Zero Product Property to set \(6w = 0\) and \(w - 5 = 0\).
- Solve the linear equations to find \(w = 0\) and \(w = 5\), which gives the final answer \(\boxed{0, 5}\).
Explanation
- Problem Analysis
We are given the quadratic equation \(6w^2 - 30w = 0\). Our goal is to find the values of \(w\) that satisfy this equation. We observe that both terms on the left side share a common factor of \(6w\).
- Factoring the Quadratic Equation
To solve the equation \(6w^2 - 30w = 0\), we factor out the greatest common factor, which is \(6w\). This gives us:
\[6w(w - 5) = 0\]
This transformation allows us to use the Zero Product Property, which states that if a product of two factors is zero, then at least one of the factors must be zero. Therefore, we have two simple cases:
- \(6w = 0\)
- \(w - 5 = 0\)
- Solving for w
Now we solve each of the two linear equations:
- For \(6w = 0\), dividing both sides by 6, we get \(w = 0\).
- For \(w - 5 = 0\), adding 5 to both sides, we get \(w = 5\).
Thus, the possible values for \(w\) that satisfy the original equation are \(0\) and \(5\).
- Conclusion
The solutions to the quadratic equation \(6w^2 - 30w = 0\) are \(w = 0\) and \(w = 5\). Thus, the values of \(w\) are \(\boxed{0, 5}\).
Examples
This type of problem, where a variable is common in all terms of a quadratic equation, often appears in physics when calculating the time at which a projectile returns to the ground. For instance, if the height of an object is given by \(h(t) = -5t^2 + 25t\), we set \(h(t) = 0\) to find the launch time (\(t=0\)) and the time it hits the ground (\(t=5\)). Understanding how to factor these equations helps in solving real-world motion problems efficiently.