Question

In Problems \(7-22,\) solve each inequality. 10\. \(x^{2}+8 x>0\)

Short answer

\((-\infty, -8) \cup (0, \, \infty) \)

Step-by-step solution

- Identify zeros of the quadratic expression

The inequality is given as \(x^2 + 8x > 0\). Start by finding the values of \(x\) that make the expression equal to 0. Set up the equation \(x^2 + 8x = 0\) and solve for \(x\).

- Factor the quadratic expression

Factor the quadratic expression from the previous step: \(x(x + 8) = 0\). This gives two solutions: \(x = 0\) and \(x = -8\).

- Determine the intervals to test

The solutions divide the number line into three intervals: \((-\text{infinity}, -8)\), \((-8, 0)\), and \((0, \text{infinity})\). Test points within these intervals to see if they satisfy the inequality \(x^2 + 8x > 0\).

- Test a point in each interval

Choose a test point in each interval: for \((-\text{infinity}, -8)\), test \(x = -9\); for \((-8, 0)\), test \(x = -1\); and for \((0, \text{infinity})\), test \(x = 1\). Substitute these values back into the inequality. For \(x = -9\), \((-9)^2 + 8(-9) = 81 - 72 = 9 > 0 \). For \(x = -1\), \((-1)^2 + 8(-1) = 1 - 8 = -7 < 0 \). For \(x = 1\), \(1^2 + 8(1) = 1 + 8 = 9 > 0\).

- Interpret the results

The inequality \(x^2 + 8x > 0\) is true for the intervals \((-\text{infinity}, -8)\) and \((0, \text{infinity})\). Conclude that the solution is \((-\infty, -8) \cup (0, \, \infty)\).

Key concepts

inequality intervals

Inequality intervals help us determine where a function is greater than or less than a certain value. In our case, we are working with the quadratic inequality given by: \[ x^2 + 8x > 0 \] To solve this, we first need to identify the values of x that make the quadratic expression equal to zero. Once the zeros are found, they divide the number line into different intervals. Here, the zeros are \[ x = 0 \text{ and } x = -8 \] These zeros divide the number line into three intervals:

• \((-\text{infinity}, -8)\)
• \((-8, 0)\)
• \((0, \text{infinity})\)

To find where the inequality is true or false, choose a test point from each interval and substitute it into the inequality to see if it satisfies the condition. For \((-\text{infinity}, -8)\), choosing \(x = -9\) gives a positive result, while \((-8, 0)\) choosing \(x = -1\) gives a negative result. Finally, choosing \(x = 1\) from \((0, \text{infinity})\) gives a positive result as well.

quadratic equation

A quadratic equation is any equation of the form \[ ax^2 + bx + c = 0 \], where \(a, b,\) and \(c\) are constants and \(x\) represents an unknown variable. For the inequality \[ x^2 + 8x > 0 \], we treat the expression like a quadratic equation \[ x^2 + 8x = 0 \] to find the critical points (roots). First, notice that the expression is already set to zero. Next, we solve the quadratic equation by factoring it, since there is no constant term \(c\). From the given inequality, we recognize that setting \[ x^2 + 8x = 0 \] allows us to find \( x(x + 8) = 0 \), giving the solutions \( x = 0 \) and \( x = -8 \). These solutions are essentially our critical points that will be used to form the inequality intervals.

factoring

Factoring is a method used to break down a complex expression into simpler parts (factors) that multiply to give the original expression. In our quadratic inequality example, we factor the polynomial \[ x^2 + 8x \]. This helps us identify the points where the quadratic expression equals zero. Factoring involves the reverse of expanding an expression. For instance, recognizing that: \[ x^2 + 8x \] can be written as \[ x(x + 8) \]. We do this by searching for a common factor in each term. Here, both \(x^2\) and \(8x\) share \(x\) as a common factor, which lets us rewrite the expression as \( x \times (x + 8) \). This factored form is easy to solve and reveals the function's critical points: \(x = 0\) and \(x = -8\). Factoring through grouping or using the quadratic formula might also sometimes be necessary if the polynomial is more complicated. However, in this specific problem, straightforward factoring suffices to simplify our inequality before further testing the intervals.