Problem 55 - Algebra 2

Question

What is the solution set for the equation \(3(4 x+1)^{2}=48 ?\) F. \(\left\\{\frac{5}{4},-\frac{3}{4}\right\\}\) G. \(\left\\{-\frac{5}{4}, \frac{3}{4}\right\\}\) H. \(\left\\{\frac{15}{4},-\frac{17}{4}\right\\}\) J. \(\left\\{\frac{1}{3},-\frac{4}{3}\right\\}\)

Step-by-Step Solution

Answer

The solution set for the equation is option G: \(-\frac{5}{4}, \frac{3}{4}\).

Step 1: Simplify the Equation

Given the equation: \[ 3(4x+1)^2 = 48 \]Begin by simplifying; divide both sides by 3 to isolate the squared term:\[ (4x+1)^2 = \frac{48}{3} = 16 \]

Step 2: Apply the Square Root

To solve for \(x\), take the square root of both sides:\[ \sqrt{(4x+1)^2} = \pm\sqrt{16} \]This gives:\[ 4x + 1 = 4 \quad \text{or} \quad 4x + 1 = -4 \]

Step 3: Solve for x (First Equation)

Solve the first equation, \(4x + 1 = 4\):Subtract 1 from both sides:\[ 4x = 3 \]Divide by 4:\[ x = \frac{3}{4} \]

Step 4: Solve for x (Second Equation)

Solve the second equation, \(4x + 1 = -4\):Subtract 1 from both sides:\[ 4x = -5 \]Divide by 4:\[ x = -\frac{5}{4} \]

Step 5: Verify the Solution Set

The solutions found are \(\frac{3}{4}\) and \(-\frac{5}{4}\). Check both values in the options:The correct solution set is \( (-\frac{5}{4}, \frac{3}{4}) \), which corresponds to option G.

Key Concepts

Algebraic Manipulation

Before diving into solving quadratic equations, it's crucial to understand algebraic manipulation. This is like reshaping an equation to make it easier to solve. In this exercise, the given equation is \(3(4x+1)^2 = 48\). Our first task is to simplify it, which means making it less complex.

By dividing both sides of the equation by 3, we managed to isolate the squared term. This is because dividing an equation by a non-zero number is a legal algebraic move that keeps the equation balanced, just like balancing a seesaw. So, we transform the equation into \((4x+1)^2 = 16\).

This is all under the umbrella of algebraic manipulation, which allows us to move forward with solving the quadratic equation.

Square Root Method

Once the equation has been simplified, the square root method becomes extremely useful. This approach is applicable when we have a perfect square on one side of the equation. In this example, our equation reaches \((4x+1)^2 = 16\).

The technique involves taking the square root of both sides of the equation. Remember, every positive real number has two square roots; one positive and one negative. This is why we write \(\sqrt{16}\) as \(\pm 4\).

Therefore, when we apply the square root to \((4x+1)^2\), we arrive at two scenarios: \(4x + 1 = 4\) and \(4x + 1 = -4\). These cases will be solved separately, leading to possibly two different roots. It’s all about exploring both paths to reach the complete solution set.

Solving Equations

With the quadratic equation simplified and the square root method applied, we now proceed to solve each resulting linear equation. Solving these equations is about finding the values of \(x\) that satisfy the equations derived from the square root method.

For \(4x + 1 = 4\), subtract 1 from both sides to get \(4x = 3\). Then, divide by 4, resulting in \(x = \frac{3}{4}\).

For the equation \(4x + 1 = -4\), follow the same steps: subtract 1 from both sides, leaving \(4x = -5\). Divide by 4, and you find \(x = -\frac{5}{4}\).

By solving both equations, we gather our solution set: \(-\frac{5}{4}\) and \(\frac{3}{4}\). Always take a moment to verify these results by substituting back into the original equation to ensure accuracy. This methodical approach covers basics from algebraic manipulation to final solution validation, helping reinforce your understanding and confidence.