Problem 54 - Precalculus Mathematics for Calculus

Question

Solve the equation by factoring. $$(2 x-1)^{2}=8$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2} + \sqrt{2}\) and \(x = \frac{1}{2} - \sqrt{2}\).

Step 1: Set Equation to Zero

First, we need to rewrite the equation so it's set equal to zero. Start with the given equation: \[(2x-1)^2 = 8\]Subtract 8 from both sides to obtain:\[(2x-1)^2 - 8 = 0\]

Step 2: Expand and Simplify

To facilitate factoring, we will expand \[(2x-1)^2\] into a quadratic expression:Use the binomial expansion formula, \((a-b)^2 = a^2 - 2ab + b^2\):\[(2x-1)^2 = (2x)^2 - 2(2x)(1) + 1^2 = 4x^2 - 4x + 1\]Replace the expanded expression in your equation:\[4x^2 - 4x + 1 - 8 = 0\]Simplify the constant terms:\[4x^2 - 4x - 7 = 0\]

Step 3: Use the Quadratic Formula

The equation \(4x^2 - 4x - 7 = 0\) doesn't factor neatly into integers, so we will use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 4\), \(b = -4\), and \(c = -7\).Plug these values into the formula:\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-7)}}{2(4)}\]Simplify inside the square root:\[x = \frac{4 \pm \sqrt{16 + 112}}{8}\]\[x = \frac{4 \pm \sqrt{128}}{8}\]Simplify the square root and fraction:\[x = \frac{4 \pm 8\sqrt{2}}{8}\]\[x = \frac{1 \pm 2\sqrt{2}}{2}\]

Step 4: Write Solution

The solutions to the equation \((2x-1)^2 = 8\), when solved by factoring (or using equivalent techniques), are:\[x = \frac{1}{2} + \sqrt{2} \quad \text{and} \quad x = \frac{1}{2} - \sqrt{2}\]

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool in solving quadratic equations that might not easily factor into integers. It's especially useful when equations can't be simplified through other factoring methods. The formula is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) represent the coefficients from the standard form of a quadratic equation, \(ax^2 + bx + c = 0\). This formula allows us to find the roots of the equation by calculating both potential solutions using the plus and minus signs in the formula.
To use it properly:

• Identify the coefficients \(a\), \(b\), and \(c\).
• Calculate the discriminant \((b^2 - 4ac)\) to determine the nature of the roots (real and distinct, real and repeated, or complex).
• Plug the values into the quadratic formula to find the solutions.

In the original problem, we used the quadratic formula because the equation \(4x^2 - 4x - 7 = 0\) did not factor neatly into integers, making it an ideal candidate for this method. This demonstrates the formula's power in tackling complex equations effectively.

Binomial Expansion

The binomial expansion technique helps in simplifying expressions raised to a power, making them easier to work with. It is derived from the binomial theorem which states:

• \((a-b)^2 = a^2 - 2ab + b^2\)

This formula is handy because it allows us to expand quadratic expressions like \((2x-1)^2\), breaking them down into simpler components.
In the provided exercise, we used the binomial expansion to expand \((2x-1)^2\) as follows:

• Square the first term \((2x)^2 = 4x^2\).
• Multiply the terms and double: \(-2(2x)(1) = -4x\).
• Square the last term \(1^2 = 1\).

So, the expanded form becomes \(4x^2 - 4x + 1\). By using binomial expansion, we were able to transform the original equation into a standard quadratic form, setting the stage for applying the quadratic formula later.

Solve Equations by Factoring

Solving equations by factoring involves expressing the equation as a product of simpler polynomials set to zero. If an equation can be factored, it allows us to find its roots more straightforwardly. Here’s how to generally approach factoring:

• Ensure the equation is set to zero.
• Look for common factors and use them to simplify the expression.
• Apply special factoring formulas if applicable (e.g., difference of squares).
• Once factored, set each factor to zero and solve for the variable.

In the exercise given, we initially attempted to solve \((2x-1)^2 = 8\) by directly factoring, but since it resulted in an unfriendly quadratic equation \(4x^2 - 4x - 7 = 0\) that didn’t factor easily into simple numbers, we opted for using the quadratic formula.
However, in many cases, factoring remains a preferred method due to its simplicity and efficiency when applicable. Understanding this method can save time and calculations, making it a must-know tool in any math enthusiast's toolkit.