Problem 46 - Basic Technical Mathematics with Calculus

Question

Solve the given problems. All numbers are accurate to at least two significant digits. Solve \(6 x^{2}-x=15\) for \(x\) by (a) factoring, (b) completing the square, and (c) the quadratic formula. Which is (a) longest?, (b) shortest?

Step-by-Step Solution

Answer

(a) Completing the square is longest, (b) Factoring is shortest. Solutions: \(x = \frac{5}{3}\) and \(x = -\frac{3}{2}\).

Step 1: Rearrange the Equation

Start by rearranging the given equation \(6x^2 - x = 15\) into standard quadratic form \(ax^2 + bx + c = 0\). Subtract 15 from both sides to get: \[6x^2 - x - 15 = 0\] Here, \(a = 6\), \(b = -1\), and \(c = -15\).

Step 2: Solve by Factoring

Attempt to factor \(6x^2 - x - 15 = 0\). We search for two numbers that multiply to \(6 \times -15 = -90\) and add to \(-1\). The numbers are \(9\) and \(-10\). Rewrite the equation as: \(6x^2 + 9x - 10x - 15 = 0\). Group terms: \((6x^2 + 9x) + (-10x - 15) = 0\). Factor each group: \(3x(2x + 3) - 5(2x + 3) = 0\). Factoring out \((2x + 3)\) gives \((3x - 5)(2x + 3) = 0\). Solve for \(x\): \(3x - 5 = 0\) yields \(x = \frac{5}{3}\), and \(2x + 3 = 0\) yields \(x = -\frac{3}{2}\).

Step 3: Solve by Completing the Square

Reorganize \(6x^2 - x - 15 = 0\) into \(6x^2 - x = 15\). Divide all terms by 6 to simplify: \(x^2 - \frac{1}{6}x = \frac{15}{6}\). To complete the square, add and subtract \(\left(\frac{1}{12}\right)^2\) to the left side: \(x^2 - \frac{1}{6}x + \left(\frac{1}{12}\right)^2 = \frac{5}{2} + \left(\frac{1}{12}\right)^2\). This forms \((x - \frac{1}{12})^2 = \frac{1801}{144}\). Taking the square root yields \(x - \frac{1}{12} = \pm \frac{\sqrt{1801}}{12}\). Solve for \(x\): \(x = \frac{1}{12} \pm \frac{\sqrt{1801}}{12}\). Calculating gives approximate values: \(x = \frac{5}{3}\) or \(x = -\frac{3}{2}\).

Step 4: Solve Using the Quadratic Formula

Use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting \(a = 6\), \(b = -1\), \(c = -15\) gives: \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(6)(-15)}}{2(6)} = \frac{1 \pm \sqrt{1 + 360}}{12}\). Simplify to \(x = \frac{1 \pm \sqrt{361}}{12}\), which results in \(x = \frac{1 \pm 19}{12}\). This gives solutions \(x = \frac{20}{12} = \frac{5}{3}\) and \(x = \frac{-18}{12} = -\frac{3}{2}\).

Step 5: Determine Longest and Shortest Methods

Analyzing the steps, the factorization method is the shortest due to direct factor identification while completing the square is the longest method due to the arithmetic gymnastics involved in transforming and solving the equation.

Key Concepts

Factoring

Factoring a quadratic equation involves expressing it as a product of two binomials. This method leverages the structure of the quadratic equation to break it down into simpler expressions. Consider the quadratic equation given:

• It starts as: \(6x^2 - x - 15 = 0\).
• The goal is to find two numbers that multiply to the product of \(a\) and \(c\), which is \(-90\), and add up to \(-1\) (the middle term coefficient).
• Here, \(9\) and \(-10\) meet the criteria because \(9 \times -10 = -90\) and \(9 - 10 = -1\).

The equation is then rewritten, grouping terms to enable factoring by grouping:

• \(6x^2 + 9x - 10x - 15 = 0\)
• Which simplifies to: \((3x(2x + 3)) - (5(2x + 3)) = 0\).

Factor out the common binomial \((2x + 3)\) leading to the factors \((3x - 5)(2x + 3) = 0\). Solving these gives the roots:\(x = \frac{5}{3}\) and \(x = -\frac{3}{2}\). Factoring is generally the quickest method, provided the equation is factorable using integers.

Completing the Square

Completing the square is a method that transforms any quadratic equation into a perfect square trinomial. This involves a series of steps to restructure the equation:

• Start with the quadratic in this form: \(6x^2 - x = 15\).
• Divide every term by 6 to simplify: \(x^2 - \frac{1}{6}x = \frac{5}{2}\).
• The magic part of this method is adding and subtracting the same number to form a perfect square trinomial.

To complete the square, calculate \((\frac{1}{12})^2\) to ensure balance. Add and subtract this on the left-hand side:

• Reformatted, the equation becomes \((x - \frac{1}{12})^2 = \frac{1801}{144}\).

The next step is to apply the square root to both sides, resulting in:

• \(x - \frac{1}{12} = \pm \frac{\sqrt{1801}}{12}\).
• Finally, solve for \(x\) to find: \(x = \frac{5}{3}\) or \(x = -\frac{3}{2}\).

This method is comprehensive and works for any quadratic equation, but it can be quite algebra-intensive and is usually the longest.

Quadratic Formula

The quadratic formula is a universal solution technique that can solve any quadratic equation in the form \(ax^2 + bx + c = 0\). By substituting the equation's coefficients into the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]You can derive solutions directly, without needing to manually factor or complete the square.

• For our equation, substitute \(a = 6\), \(b = -1\), and \(c = -15\) into the formula.
• Calculate the discriminant: \(b^2 - 4ac = 1 + 360 = 361\).
• The solutions then compute as: \(x = \frac{-(-1) \pm \sqrt{361}}{12} = \frac{1 \pm 19}{12}\), yielding \(x = \frac{5}{3}\) and \(x = -\frac{3}{2}\).

The quadratic formula provides a straightforward manner to find roots and is preferred if factoring is not straightforward or completing the square seems cumbersome.