Question

Solve the inequality: \(x^{2}>x\).

Short answer

The solution to the inequality \(x^2 > x\) is \(x \in (-\infty, 0) \cup (1, \infty)\).

Step-by-step solution

Rewrite the inequality and form a quadratic equation

To start solving the inequality, rewrite the given inequality by moving all terms to one side. This will help us to form a quadratic equation. Given inequality: \(x^2 > x\) Rewrite the inequality as: \(x^2 - x > 0\) Now, we will identify the critical points by finding the roots of the quadratic equation \(x^2 - x = 0\).

Find the critical points

To find the critical points, we will factor the quadratic equation and solve for x. \(x^2 - x = 0\) Factor out x: \(x(x - 1) = 0\) Now, we can find the roots of the equation: For \(x(x - 1) = 0\), either \(x = 0\) or \(x - 1 = 0\) Thus, the critical points are: \(x = 0\) and \(x = 1\)

Determine the sign of the inequality in each interval between the critical points

Now that we have our critical points, let's determine the sign of the inequality in each interval between the critical points. There are three intervals to consider: 1. \(x < 0\) 2. \(0 < x < 1\) 3. \(x > 1\) Let's check the sign of the inequality in each interval. Interval 1: \(x < 0\) For a negative x, both factors (x) and (x-1) are negative: Negative * Negative > 0 (True) The inequality holds true for this interval. Interval 2: \(0 < x < 1\) In this interval, the factor x is positive, and the factor (x-1) is negative: Positive * Negative > 0 (False) The inequality does not hold true for this interval. Interval 3: \(x > 1\) For x > 1, both factors x and (x-1) are positive: Positive * Positive > 0 (True) The inequality holds true for this interval.

Write the solution for the inequality

Based on our findings, the inequality \(x^2 - x > 0\) is true for intervals 1 and 3 (i.e., when \(x < 0\) and when \(x > 1\)). Hence, the solution of the inequality is: \(x \in (-\infty, 0) \cup (1, \infty)\)

Key concepts

Quadratic Equation

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. Solving quadratic equations is a staple in algebra and helps in finding the roots or solutions where the expression equals zero.

In our exercise, by transforming the inequality \(x^2 > x\) into the quadratic equation \(x^2 - x = 0\), we create an expression that highlights the boundary points or critical points where the value changes from greater than zero to less than zero or vice versa.

Quadratic equations can usually be solved by factoring, completing the square, or using the quadratic formula. Here, our example was simple enough to factor because it can be written as \(x(x - 1) = 0\). Recognizing quadratic forms helps us identify potential solutions efficiently.

Critical Points

Critical points are specific values of \(x\) where the behavior of a function or inequality changes. In our case, these are the points where the quadratic equation \(x^2 - x = 0\) equals zero. Finding these points is crucial in solving inequalities.

To identify critical points, we set the quadratic to zero, \(x(x - 1) = 0\), and solve for \(x\). This yields two solutions: \(x = 0\) and \(x = 1\). These values divide the number line into intervals that we will analyze further.

• \(x = 0\): This is one of the roots where the factor \(x\) equals zero.
• \(x = 1\): Here, the factor \(x-1\) becomes zero.

Understanding these changes helps in determining which intervals satisfy the original inequality. Critical points don't always satisfy the inequality themselves, but they do mark changes in the inequality pattern.

Number Line Analysis

Number line analysis is a method used to evaluate inequalities by considering different intervals defined by the critical points. With our critical points \(x = 0\) and \(x = 1\), the number line is divided into three intervals: \((-fty, 0)\), \((0, 1)\), and \((1, fty)\). Analyzing these intervals helps us see where the inequality holds.

For each interval, we pick a test point and substitute it into the inequality \(x^2 - x > 0\):

• **Interval \(x < 0\)**: Choose \(x = -1\). Substituting gives \((-1)^2 - (-1) = 1 + 1 > 0\), which is true.
• **Interval \(0 < x < 1\)**: Choose \(x = 0.5\). Substituting gives \(0.5^2 - 0.5 = 0.25 - 0.5 = -0.25 < 0\), which is false.
• **Interval \(x > 1\)**: Choose \(x = 2\). Substituting gives \(2^2 - 2 = 4 - 2 > 0\), which is true.

This analysis confirms where the original inequality is satisfied, leading to our solution: \(x \in (-fty, 0) \cup (1, fty)\). This technique ensures accuracy by visually representing the solutions.