Solve each quadratic inequality. Use interval notation to write each solution set. 12 x^2+8 x=15

Name: Problem 58, Chapter 1: Essentials of Precalculus
Uploaded: 2021-07-09T20:49:24-08:00
Duration: 4 min 5 s
Channel: Julie Silva
Description: Solve each quadratic inequality. Use interval notation to write each solution set. 12 x^2+8 x=15

step 1 We're being asked to solve the given quadratic inequality. So first, we have to set it equal to

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

Key Concepts

Quadratic Inequalities

A quadratic inequality involves a quadratic expression set in relation to zero (or another expression) with inequality symbols like <, >, ?, or ?. To solve these, one first rewrites the inequality in standard form, identifies the roots, and then determines sign changes across intervals to decide which regions satisfy the inequality.

Quadratic Equations

To solve quadratic inequalities, one often begins by solving the corresponding quadratic equation (obtained by replacing the inequality sign with an equals sign). This step isolates the critical points or roots of the quadratic, which serve as boundaries that divide the number line into intervals for further testing.

Critical Points and Interval Testing

Once the roots of the quadratic equation are found, these points divide the number line into several intervals. By choosing test points from each interval and evaluating the original quadratic expression, one can determine whether the expression is positive or negative in that region and hence which intervals satisfy the inequality.

Interval Notation

Interval notation is a concise way to represent sets of numbers on the number line, using parentheses to indicate open endpoints (where the endpoint is not included) and brackets for closed endpoints (where the endpoint is included). After determining the intervals that satisfy the inequality, the solution set is expressed using this notation.

Transcript

00:01 We're being asked to solve the given quadratic inequality.

00:03 So first, we have to set it equal to zero.

00:05 So i'll subtract 15 from both sides.

00:08 So that will give us 12x squared plus 8x minus 15 is greater than they're equal to zero.

00:15 Perfect.

00:16 So the next thing we need to do is find the zeros, meaning we're going to set 12x squared plus 8x minus 15 equal to zero.

00:23 And i'm going to solve this by factoring.

00:26 Well, there is no greatest common factor.

00:27 And because our leading coefficient is greater than one, i'm going to go a grouping method.

00:32 So i'll begin by multiplying the first and last coefficients.

00:36 Well, 12 times negative 15 is equal to negative 180.

00:40 So now we have to find two numbers that multiply the negative 180 that will add to 8.

00:45 So i'm thinking of 18 and negative 10.

00:48 18 plus negative 10 is 8.

00:51 18 times negative 10 is negative 180.

00:53 Perfect.

00:55 So now i'm going to rewrite my problem, but in place of a.

00:58 I'm going to substitute in 18x minus 10x.

01:01 And we'll bring down our first term and our last term.

01:06 And now we have four terms, so we'll factor by grouping.

01:09 So we'll split the problem in half and factor out the greatest common factor from the first two terms, which is 6x.

01:15 So when we divide both terms by 6x, we're left with 2x plus 3.

01:20 Then we'll factor out the greatest common factor from the second two terms, which is negative 5.

01:26 And when we divide both terms by negative 5, we're also left with 2x plus 3.

01:30 So as you can see, they both have this common factor of 2x plus 3.

01:35 So that's our first factor.

01:37 And our second factor are the two terms on the outside, 6x minus 5.

01:42 So now we just have to set both factors equal to 0.

01:46 So 2x plus 3 could equal to 0 or 6x minus 5 could equal to 0.

01:51 And now we solve both equations.

01:53 So to solve the first equation, we'll subtract 3 from both sides.

01:56 So 2x will equal to negative 3.

01:58 Then we'll divide both sides by 2...

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