Solve each quadratic inequality. Use interval notation to write each solution set. x^2<-x+30

Name: Problem 56, Chapter 1: Essentials of Precalculus
Uploaded: 2021-07-09T20:39:58-08:00
Duration: 2 min 25 s
Channel: Julie Silva
Description: Solve each quadratic inequality. Use interval notation to write each solution set. x^2<-x+30

Solve each quadratic inequality. Use interval notation to...

Key Concepts

Interval Notation

Interval notation is a concise method to denote sets of numbers that satisfy an inequality. It uses parentheses and brackets to indicate open and closed intervals, respectively, providing a clear, standardized way to express the solution set of a quadratic inequality.

Sign Analysis

Sign analysis entails determining whether the quadratic expression is positive or negative within intervals separated by its zeros. By testing points within each interval, one can decide which regions satisfy the inequality, based on the behavior of the parabola (whether it opens upward or downward).

Finding Zeros

The zeros or roots of the quadratic expression are critical because they are the points where the expression equals zero and typically mark the boundaries between regions of different signs. These can be found through factoring, the quadratic formula, or completing the square, and they divide the number line into intervals for further analysis.

Standard Form

Rewriting a quadratic inequality into standard form (usually ax² + bx + c < 0 or > 0) is an important step. This form makes it easier to analyze and solve because it consolidates all terms on one side of the inequality, thereby establishing a clear threshold (zero) for determining where the expression changes sign.

Quadratic Inequalities

A quadratic inequality involves a quadratic expression compared to zero (or another expression) using inequality symbols. To solve such inequalities, one typically rewrites the expression in a standard form and identifies the regions where the quadratic expression is positive or negative, often based on the graph of a parabola.

Transcript

00:01 In this problem, we're being asked to solve the given quadratic inequality.

00:04 Well, first, we need to set it to zero.

00:06 So i'm going to start by adding x to both sides, and at the same time, i'm going to subtract 30.

00:12 So that will leave us with x squared plus x minus 30 is less than zero.

00:18 Perfect.

00:19 So now that it sets a zero, we now need to find the zeros for our quadratic, meaning we have to set our quadratic equation equal to zero.

00:27 And i'm going to solve this by factoring.

00:29 Well, there is no greatest common factor, and because the coefficient of our first term is one, we can set up both binomials and know that they both start with x because x times x is x squared.

00:40 So now we just have to find two integers that multiply the negative 30 that add to our middle coefficient of positive 1.

00:48 So i'm thinking positive 6 and negative 5.

00:51 6 plus negative 5 is 1, 6 times negative 5 is negative 30.

00:56 So we'll substitute those into our factors.

00:59 So now that we have our factors to solve, we're going to set both of them equal to zero and then solve the equations.

01:05 So to solve the first, we subtract six from both sides, so x would be negative six.

01:10 And to solve the second, we add five to both sides, so x could be equal to five.

01:14 Perfect.

01:16 So the next thing we need to do is set up a sign chart...

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