Problem 69 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$(5 x+3)(5 x-3)-10(x-1)=0$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{5}\).

Step 1: Expand the Expression

First, we need to expand the left side of the equation. Notice that \[(5x + 3)(5x - 3)\]is a difference of squares. It simplifies to \[(5x)^2 - (3)^2 = 25x^2 - 9.\]So the equation becomes:\[25x^2 - 9 - 10(x - 1) = 0.\]

Step 2: Distribute Terms Inside the Parentheses

Next, distribute \(-10\) across the terms inside the parentheses, \(x - 1\):\[-10(x - 1) = -10x + 10.\]Substitute this back into the equation:\[25x^2 - 9 - 10x + 10 = 0.\]

Step 3: Simplify the Equation

Combine the constant terms in the equation:\[25x^2 - 10x + 1 = 0.\]

Step 4: Identify Coefficients for Quadratic Formula

The equation \(25x^2 - 10x + 1 = 0\) is in the standard quadratic form \(ax^2 + bx + c = 0\), where \(a = 25\), \(b = -10\), and \(c = 1\).

Step 5: Apply the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute in the identified coefficients:\[x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 25 \cdot 1}}{2 \cdot 25}.\]This simplifies to:\[x = \frac{10 \pm \sqrt{100 - 100}}{50}.\]

Step 6: Simplify the Square Root

Since \(100 - 100 = 0\), the square root reduces to:\[\sqrt{0} = 0.\]So the expression becomes:\[x = \frac{10 \pm 0}{50}.\]

Step 7: Solve for x

With the expression simplified, solve for \(x\):\[x = \frac{10}{50} = \frac{1}{5}.\]Since there is no \(\pm\) term remaining after simplification, there is only one solution.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). The formula helps to find the roots (solutions) of these equations by using the coefficients \( a \), \( b \), and \( c \). The quadratic formula is:

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

Bringing the quadratic equation to this standard form is crucial. Once you have identified the coefficients, plug them into the formula, carefully solving the operations under the square root and the arithmetic to find the roots. In the problem above, using the quadratic formula ultimately zeroes out the discriminant (\( b^2 - 4ac \)) to yield a single solution because the square root of zero is zero.

Difference of Squares

The difference of squares is a significant concept in algebra, especially when working with expressions like \((5x + 3)(5x - 3)\). This expression looks complex, but it simplifies nicely due to the nature of the difference of squares. When you have a product like this, it takes the form:

• \((a + b)(a - b) = a^2 - b^2\)

In the exercise, \( (5x)^2 - (3)^2 = 25x^2 - 9 \) is precisely the result of applying this identity. Recognizing these patterns makes it easier to expand expressions efficiently, reducing errors in complex calculations. It's a shortcut that transforms apparent complexity into simplicity.

Simplifying Equations

Simplifying equations is a vital skill in solving mathematical problems. Always aim to reduce the complexity of equations by following a systematic process of simplification. For the equation we worked with, initially write the expanded form, then methodically combine like terms and rearrange all components into a simpler structure. The goal is to have an easily manageable form where you can clearly identify coefficients for further steps like applying the quadratic formula. Simplification might involve distributing terms, like how \(-10(x - 1)\) becomes \(-10x + 10\), and then aligning the equation to a clear \( ax^2 + bx + c = 0 \) format.

Identifying Coefficients

Identifying coefficients is critical to solving quadratic equations. A quadratic equation must be in the standard form \( ax^2 + bx + c = 0 \) before you can use specific methods like factoring or the quadratic formula. In the exercise, the equation \( 25x^2 - 10x + 1 = 0 \) is a clear example.Here:

• \( a = 25 \)
• \( b = -10 \)
• \( c = 1 \)

Accurately identifying these coefficients ensures proper application of the quadratic formula. Each plays a distinct role in the calculation process. The value of \( a \) influences the "width" of the parabola the equation represents, \( b \) affects the symmetry, and \( c \) determines where the parabola intersects the vertical axis.