Problem 57 - Precalculus Mathematics for Calculus

Question

Find all real solutions of the equation, rounded to two decimals. $$x(x-1)(x+2)=\frac{1}{6} x$$

Step-by-Step Solution

Answer

Real solutions are \(x = 0\), \(x = 1.06\), and \(x = -2.06\).

Step 1: Set Equation to Zero

First, let's rearrange the original equation to set it equal to zero. Start by subtracting \(\frac{1}{6}x\) from both sides:\[x(x-1)(x+2) - \frac{1}{6}x = 0\]

Step 2: Distribute and Simplify the Polynomial

Expand the left side first:\[x(x^2 + x - 2)\]This becomes:\[x^3 + x^2 - 2x\]Now, subtract \(\frac{1}{6}x\) from the equation:\[x^3 + x^2 - 2x - \frac{1}{6}x = 0\]Factor out \(x\) from the equation:\[x \left( x^2 + x - 2 - \frac{1}{6} \right) = 0\]

Step 3: Convert to a Common Denominator

Combine like terms in the quadratic expression by converting \(\frac{1}{6}\) to a common denominator:\[x^2 + x - \frac{12}{6} - \frac{1}{6} = x^2 + x - \frac{13}{6}\]Thus, the equation becomes:\[x \left( x^2 + x - \frac{13}{6} \right) = 0\]

Step 4: Find Solutions from Factored Form

Now solve for \(x\):1. \(x = 0\)2. For the quadratic equation:\[x^2 + x - \frac{13}{6} = 0\]use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute \(a = 1\), \(b = 1\), \(c = -\frac{13}{6}\)\[x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times \left(-\frac{13}{6}\right)}}{2 \times 1}\]

Step 5: Simplify the Quadratic Formula Solution

Calculate the discriminant and solve:\[x = \frac{-1 \pm \sqrt{1 + \frac{52}{6}}}{2} = \frac{-1 \pm \sqrt{\frac{58}{6}}}{2}\]\[x = \frac{-1 \pm \sqrt{9.67}}{2}\]Thus:1. \(x = \frac{-1 + 3.11}{2} = 1.06\) (rounded)2. \(x = \frac{-1 - 3.11}{2} = -2.06\) (rounded)So, the solutions are \(x = 0\), \(x = 1.06\), and \(x = -2.06\).

Key Concepts

Real Solutions

Real solutions, in the context of polynomial equations like the one we are solving, refer to the values of the variable that make the equation true. These are the points where the graph of the polynomial intersects the x-axis. In simpler terms, they are the x-values where the polynomial equals zero.

When solving equations, particularly polynomials, the goal is usually to find all real solutions. This means finding all x-values that satisfy the equation. These are often called 'roots' or 'zeroes'. For a real solution to exist, the calculation under the square root sign in the quadratic formula, known as the discriminant, should be non-negative. If the discriminant is positive, you get two real solutions; if it's zero, you get one; and if it's negative, there are no real solutions.

In our case, solutions include both straightforward values, like 0, and those calculated using methods like the quadratic formula. It's crucial to ensure all resulting values are precise up to the specified decimal place, ensuring accuracy in solving real-world problems.

Quadratic Formula

The quadratic formula is a mathematical formula used for finding the solutions of a quadratic equation, which is any equation that can be rewritten in the form \(ax^2 + bx + c = 0\). The quadratic formula is:

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

This simple yet powerful tool allows you to find solutions for any quadratic equation by substituting the appropriate values for \(a\), \(b\), and \(c\) into the formula. It's especially useful when the quadratic doesn't factorise neatly or when trying to find decimal solutions.

In our example, after setting up the equation correctly, we have \(a = 1\), \(b = 1\), and \(c = -\frac{13}{6}\). Plugging these values into the quadratic formula helps us find the real solutions. The discriminant \(b^2 - 4ac\) guides us on the number of solutions, and if real solutions exist. Here, it leads us to calculate the two additional solutions, 1.06 and -2.06, adding these to our previously found whole number solution, 0.

Understanding and applying the quadratic formula is vital because it offers a systematic approach to solving any quadratic equation, ensuring you find all possible solutions.

Factoring Polynomial Expressions

Factoring polynomial expressions involves breaking down a polynomial into simpler terms, or factors, that when multiplied together give you the original polynomial. This is a fundamental algebraic technique useful for solving polynomial equations since it allows equations to be simplified and made easier to solve.

In our task, the original polynomial expression \(x(x-1)(x+2)\) is transformed and simplified through expansion and simplification. After subtracting \(\frac{1}{6}x\), further simplification involves factoring out common terms, such as \(x\), transforming the expression into \(x \left( x^2 + x - \frac{13}{6} \right) = 0\).

• The first factor, \(x = 0\), provides a straightforward solution.
• The second factor leads us to use the quadratic formula for further solutions.

Factoring assists in identifying the solutions or steps towards solutions for polynomial equations. It's often the preliminary step before applying other methods like the quadratic formula, especially when some terms can be factored out easily, simplifying the process of finding real solutions.