Problem 28 - Elementary Algebra

Question

The area of a rectangle is given by the function \(A(x)=10 x 2-59 x-6\), where \(x\) is measured in meters. Rewrite this function in factored form.

Step-by-Step Solution

Answer

The factored form is \((10x + 1)(x - 6)\).

Step 1: Understand the problem

The given problem involves rewriting the quadratic function for the area of a rectangle in factored form. The original function is given as \( A(x) = 10x^2 - 59x - 6 \). We want to express this quadratic expression as a product of two binomials.

Step 2: Identify coefficients

The quadratic function is in the form \( ax^2 + bx + c \), where \( a = 10 \), \( b = -59 \), and \( c = -6 \). Factor the quadratic expression using these coefficients.

Step 3: Factor using the quadratic formula

To factor the quadratic expression, we can begin by finding the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substituting the values of \( a \), \( b \), and \( c \) into the formula gives us: \[ x = \frac{-(-59) \pm \sqrt{(-59)^2 - 4 \cdot 10 \cdot (-6)}}{2 \cdot 10} \]

Step 4: Simplify the quadratic formula

Calculating inside the square root: \((-59)^2 = 3481\) and \(4 \cdot 10 \cdot (-6) = -240\). Thus, \(3481 + 240 = 3721\). Now, calculate the roots: \[ x = \frac{59 \pm \sqrt{3721}}{20} \] Since \(\sqrt{3721} = 61\), substituting back gives: \[ x = \frac{59 \pm 61}{20} \]

Step 5: Calculate the roots

Two cases arise from the expression: \( x = \frac{59 + 61}{20} = 6 \) and \( x = \frac{59 - 61}{20} = -\frac{1}{10} \). So, the two roots of the quadratic function are \( x = 6 \) and \( x = -\frac{1}{10} \).

Step 6: Write the factored form

Using the roots, the factored form of the quadratic is \((x - 6)\left(x + \frac{1}{10}\right)\). Scaling the entire expression to eliminate fractions gives \((10x + 1)(x - 6)\). Expand this to check against the original function as a verification step.

Step 7: Verification

Expand \((10x + 1)(x - 6)\) to ensure it gives the original quadratic function. Calculating: \(10x^2 - 60x + x - 6 = 10x^2 - 59x - 6\), which verifies this is correct.

Key Concepts

Quadratic Formula

The quadratic formula is a universal tool used to find the roots or solutions of any quadratic equation. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). The quadratic formula is:

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

To use this formula effectively, it is important to identify the coefficients \(a\), \(b\), and \(c\) from your specific equation. In the context of our exercise, they were 10, -59, and -6, respectively.
By substituting these values into the formula, we found the roots of the function \( A(x) = 10x^2 - 59x - 6 \).
It is crucial to carefully execute the arithmetic and ensure the calculations under the square root (the discriminant) are accurate. In this case, calculating \( b^2 - 4ac \) gave us a discriminant of 3721, leading to real and rational roots.

Roots of Quadratic Equations

The roots of a quadratic equation are the values of \(x\) that make the equation equal to zero. They are also known as zeros or solutions of the equation. In our exercise, we found the roots by solving:

• \( x = \frac{59 \pm 61}{20} \)

This results in the values \( x = 6 \) and \( x = -\frac{1}{10} \).
By substituting these values back into the original equation, both should result in the function being equal to zero, confirming their validity as roots.
These roots also have a geometric interpretation: they represent the points where the parabola (graph of the quadratic function) intersects the x-axis.

Factored Form

Rewriting a quadratic function in factored form involves expressing it as a product of two binomials. This forms:

• \( A(x) = (x - \text{root}_1)(x - \text{root}_2) \)

From our exercise, the factored form of \( A(x) = 10x^2 - 59x - 6 \) is \((10x + 1)(x - 6)\).
We found this form by using the roots \( x = 6 \) and \( x = -\frac{1}{10} \).
Understanding factoring is important because it allows for simplification of expressions and solving equations more easily. Always verify by expanding back to ensure it yields the original quadratic expression, confirming it's correct.