Problem 44 - Algebra 2

Question

Solve each equation by using the Square Root Property. \(4 x^{2}-28 x+49=5\)

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = 5\).

Step 1: Rewrite the Equation

Begin by rewriting the given equation in a form that allows for easy application of the Square Root Property. We have:\[4x^2 - 28x + 49 = 5\]Subtract 5 from both sides to set the equation to zero:\[4x^2 - 28x + 49 - 5 = 0\]This simplifies to:\[4x^2 - 28x + 44 = 0\]

Step 2: Factor the Equation

Notice that the left side of the equation resembles a perfect square. Factor it as:\[(2x - 7)^2 = 9\]This is because \( (2x - 7)^2 = 4x^2 - 28x + 49 \), and the equation becomes:\[(2x - 7)^2 = 9\]

Step 3: Apply the Square Root Property

The Square Root Property states that if \((a)^2 = b\), then \(a = \pm \sqrt{b}\). Apply this property to solve for \(x\):\[2x - 7 = \pm 3\]This gives us two possible equations: \[2x - 7 = 3\]\[2x - 7 = -3\]

Step 4: Solve Each Equation

Solve the two equations separately.For \(2x - 7 = 3\):1. Add 7 to both sides: \[2x = 10\]2. Divide by 2: \[x = 5\]For \(2x - 7 = -3\):1. Add 7 to both sides: \[2x = 4\]2. Divide by 2: \[x = 2\]

Step 5: Write the Solution Set

The solutions to the equation \(4x^2 - 28x + 49 = 5\) are the values of \(x\) that satisfy either of the equations solved in the previous step.Thus, the solution set is:\[x = 2, x = 5\]

Key Concepts

Solving Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They are central in algebra and have various methods for solving. One powerful method is the Square Root Property. This property simplifies solving by focusing on equations expressed as perfect squares.
When dealing with a quadratic equation, the ultimate goal is to rewrite it in a way that makes taking the square root possible. This usually means getting the equation to look like \( (x - p)^2 = q \) where you can find \(x\) by using the Square Root Property:

• If \( (a)^2 = b \), then \( a = \pm \sqrt{b} \).

This method significantly reduces the complexity when compared to solving using factoring or the quadratic formula, especially when the expression is a perfect square.

Factoring Perfect Squares

Factoring perfect squares is a method used to simplify quadratic equations by recognizing specific patterns. In this exercise, the quadratic expression \(4x^2 - 28x + 49\) can be rewritten as \((2x - 7)^2\).

If a quadratic equation can be expressed in the form \((ax - b)^2\), it's considered a perfect square trinomial. Identifying these can drastically simplify solving the equation using the Square Root Property.

• The perfect square trinomial pattern follows \( (a-b)^2 = a^2 - 2ab + b^2 \).
• Recognizing the pattern in \( 4x^2 - 28x + 49 = (2x - 7)^2 \) helps us to factor quickly.

Once we've factored into a perfect square, the equation becomes much simpler to work with, especially for applying further properties like the Square Root Property.

Solution Sets

The solution set of an equation consists of all the values that satisfy the equation. In this case, we started with the quadratic equation \(4x^2 - 28x + 44 = 0\), after rewriting and factoring, we used the Square Root Property.

By setting \( (2x - 7)^2 = 9 \), we found two solutions:

• Solving \( 2x - 7 = 3 \) gives \(x = 5\).
• Solving \( 2x - 7 = -3 \) gives \(x = 2\).

The solution set is simply the collection of all solutions, noted as \(\{2, 5\}\). This means that substituting \(x = 2\) or \(x = 5\) will satisfy the original equation \(4x^2 - 28x + 49 = 5\).
Recognizing and expressing the solution set clearly allows you to understand all possible answers to the equation, ensuring that no potential solution is overlooked.