What are the solutions of the quadratic equation $0=4(x-3)^2-16$ ?
The problem asks us to solve the quadratic equation \(0 = 4(x-3)^2 - 16\) for the variable \(x\). We have a quadratic expression set to zero, which we can solve by isolating the squared term and then taking the square root. The equation is given in a form that suggests using the square root method, which is often more efficient than expanding the binomial.
To solve for \(x\), we first isolate the squared binomial term. Starting with the equation:
\[4(x-3)^2 - 16 = 0\]
Adding 16 to both sides gives:
\[4(x-3)^2 = 16\]
Next, divide both sides by 4 to isolate \((x-3)^2\):
\[(x-3)^2 = \frac{16}{4}\]
\[(x-3)^2 = 4\]
Now, take the square root of both sides. Remember that taking the square root of a positive constant results in both a positive and a negative root:
\[x - 3 = \pm \sqrt{4}\]
\[x - 3 = \pm 2\]
This gives us two separate linear equations:
Finally, solve each of the two resulting equations for \(x\):
\(x = 2 + 3 = 5\)
\(x = -2 + 3 = 1\)
The solutions to the quadratic equation are \(x = 5\) and \(x = 1\). Comparing these to the provided options, the correct pair is \(1\) and \(5\).
Quadratic equations like this are very common in physics and engineering. For example, if you are calculating the time it takes for an object thrown into the air to reach a certain height, the path of the object can be modeled by a quadratic equation. Solving for the time variable \(t\) involves similar steps to the ones we used here, helping determine exactly when an event occurs, such as when an object lands or reaches a target.