Question

What is the solution set to the equation \(0=(3 a+1)^2(a-4)\) ? A) \(\left\\{\frac{1}{3},-4\right\\}\) ?) \(\left\\{-\frac{1}{3}, 4\right\\}\) C) \(\left\\{-\frac{1}{3}, \frac{1}{3},-4\right\\}\) D) \(\left\\{-\frac{1}{3}, \frac{1}{3}, 4\right\\}\)

Short answer

The solution set to the equation \(0=(3 a+1)^2(a-4)\) is B) \(\left\{-\frac{1}{3}, 4\right\}\).

Step-by-step solution

Identify the equation

The given equation is: \(0 = (3a + 1)^2(a - 4)\)

Utilize the zero-product property

According to the zero-product property, if the product of two or more factors is zero, then at least one of the factors must be zero. In this case, we have two factors: \((3a + 1)^2\) and \((a - 4)\). So, either \((3a + 1)^2 = 0\) or \((a - 4) = 0\).

Solve for \(a\) in the first factor

To solve the equation \((3a + 1)^2 = 0\), take the square root of both sides: \[\sqrt{(3a + 1)^2} = \sqrt{0}\] \[3a + 1 = 0\] Now, solve for \(a\): \[3a = -1\] \[a = -\frac{1}{3}\]

Solve for \(a\) in the second factor

To solve the equation \((a - 4) = 0\), simply add 4 to both sides: \[a = 4\]

Write the solution set

The values of \(a\) that satisfy the equation are \(-\frac{1}{3}\) and \(4\). So, the solution set is: \(\left\\{-\frac{1}{3}, 4\right\\}\) The correct answer is option B) \(\left\\{-\frac{1}{3}, 4\right\\}\).