Problem 10 - College Algebra

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ x^{2}-2 x+1>0 $$

Step-by-Step Solution

Answer

The solution set for the given inequality, in interval notation, is \((-\infty, 1) \cup (1, +\infty)\).

Step 1: Factoring the Quadratic Polynomial

A quadratic polynomial \(x^{2}-2 x+1\) can be factored as \((x-1)^{2}\)

Step 2: Solving the Inequality

Find the roots of the inequality by setting \((x-1)^{2}>0\). This gives one valid solution, \(x=1\)

Step 3: Check Intervals

Divide the number line into intervals using the root and test an element from each interval to see whether it satisfies the inequality. Since the inequality is a perfect square, it is always positive (i.e., \((x-1)^{2}\geq0\) for all real numbers x), except at x = 1, where it equals 0.

Step 4: Write the Solution in Interval Notation

Because the original inequality is strict (\(>\), not \(\geq\)), the solution does not include x = 1. Therefore, the solution in interval notation is \((-\infty, 1) \cup (1, +\infty)\).

Key Concepts

Factoring Quadratic Polynomials

Understanding how to factor quadratic polynomials is essential in solving polynomial inequalities. Factoring simplifies the polynomial into a product of binomials or simpler polynomials, which makes finding the roots easier.

Consider a quadratic polynomial in the form of \( ax^2 + bx + c \). One common method of factoring involves looking for two numbers that multiply to \( ac \) and add up to \( b \). Once these numbers are found, they are used to split the middle term and factor by grouping, resulting in two binomials.

In the exercise \( x^2 - 2x + 1 \), factoring is straightforward since it's a perfect square trinomial—where the first and last terms are perfect squares, and the middle term is twice the product of the square roots of the first and last terms. This trinomial neatly factors into \( (x - 1)^2 \), illustrating how recognizing special patterns can make the process quicker and more efficient.

Interval Notation

Interval notation is a method of writing sets of numbers that satisfy a particular condition. It's an efficient way to express continuous ranges where inequalities are present. In interval notation, we use brackets and parentheses to denote closed and open intervals, respectively.

For instance, a closed interval \( [a, b] \) includes the endpoints \( a \) and \( b \), meaning all numbers between and including \( a \) and \( b \) satisfy the condition. An open interval \( (a, b) \), on the other hand, does not include the endpoints, representing all numbers strictly between \( a \) and \( b \).

When an interval stretches infinitely in one direction, we use the symbols \( -\text{\(\fy{\text{ty}}\text{y}{3}\f-1in)} \)fty) \) or \( +\text{\(\fy{\text{ty}}\)} \) to indicate no bound in that direction. Therefore, for a solution such as \( x > a \), we would write \( (a, +\text{\(\fy{\text{ty}}\)} ) \). Always remember, infinity symbols always get parentheses, as infinity itself is not a number that can be reached.

Inequality Graphing

Graphing inequalities on a number line helps visualize the solution set of an inequality. It's a simple yet powerful way to convey which numbers satisfy the inequality.

To graph an inequality, one usually starts by marking the roots of the corresponding equality on the number line. In the case of \( (x-1)^2 > 0 \), we mark the root at \( x = 1 \). Then, we test points in each of the regions divided by the roots to determine where the inequality holds true.

For strict inequalities (\( > \) or \( < \)), we use open circles to indicate that the roots themselves are not part of the solution set. Similarly, for non-strict inequalities (\( \f-\text{\fy{}{2}\f-1-1}geq\) or \(\f-\text{\fy{}{2}\f-1-1}{leq} \)), we use closed circles to include the roots. After testing the intervals, we shade the regions of the number line where the inequality is true. In this exercise, since \( (x-1)^2 \) is positive except at \( x = 1 \), the solution consists of all other numbers, resulting in two open intervals extending indefinitely in both directions from the root.