Title

College Algebra

WTAMU > Virtual Math Lab > College Algebra > Tutorial 23A: Quadratic Inequalities

 Answer/Discussion to 1a I used the test-point method, but you could also use the sign graph of factors method.

 Step 1: Write the quadratic inequality in standard form.

 *Inv. of sub. 5x is add. 5x

 Step 2: Solve the quadratic equation, , to get the boundary point(s).

 *Factor   *Set 1st factor = 0 and solve             *Set 2nd factor = 0 and solve

 -6 and 1 are boundary points.

 Below is a graph that marks off the boundary points -6 and 1 and shows the three sections that those points have created on the graph.  Note that open holes were used on those two points since our original inequality does not include where it is equal to 0.  Note that the two boundary points create three sections on the graph: , , and .

 Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.

 You can choose ANY point in an interval to represent that interval.  Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. Keep in mind that our inequality is .  Since we are looking for the quadratic expression to be GREATER THAN 0, that means we need our sign to be POSITIVE. From the interval , I choose to use -7 to test this interval: (I could have used -10, -25, or -10000 as long as it is in the interval)

 *Chose -7 from 1st interval to plug in for x

 Since 8 is positive and we are looking for values that cause our quadratic expression to be greater than 0 (positive),  would be part of the solution.

 From the interval , I choose to use 0 to test this interval. (I could have used -5, -4, or -3 as long as it is in the interval)

 *Chose 0 from 2nd interval to plug in for x

 Since -6 is negative and we are looking for values that cause our expression to be greater than 0 (positive),  would not be part of the solution.

 From the interval , I choose to use 2 to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval)

 *Chose 2 from 3rd interval to plug in for x

 Since 8 is positive and we are looking for values that cause our quadratic expression to be greater than 0 (positive),  would be part of the solution.

 Interval notation:  Graph: *Open intervals indicating all values less than -6 or greater than 1 *Visual showing all numbers less than -6 or greater than 1

 Answer/Discussion to 1b I used the test-point method, but you could also use the sign graph of factors method.

 Step 1: Write the quadratic inequality in standard form.

 Step 2: Solve the quadratic equation, , to get the boundary point(s).

 *Factor *Set 1st factor = 0 and solve                     *Set 2nd factor = 0 and solve

 1/3 and 1/2 are boundary points.

 Below is a graph that marks off the boundary points, 1/3 and 1/2, and shows the three sections that those points have created on the graph.  Note that closed holes were used on those two points since our original inequality includes where it is equal to 0.  Note that the two boundary points create three sections on the graph: ,  , and .

 Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.

 You can choose ANY point in an interval to represent that interval.  Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get. Keep in mind that our original problem is .  Since we are looking for the quadratic expression to be LESS THAN OR EQUAL TO 0, that means we need our sign to be NEGATIVE (OR 0). From the interval , I choose to use 0 to test this interval: (I could have used -10, -25, or -10000 as long as it is in the interval)

 *Chose 0 from 1st interval to plug in for x

 Since 1 is positive and we are looking for values that cause our quadratic expression to be less than or equal to 0 (negative or 0), would not be part of the solution.

 From the interval , I choose to use 2/5 to test this interval. (I could have used 5/12 or 3/8 as long as it is in the interval)

 *Chose 2/5 from 2nd interval to plug in for x

 Since -1/25 is negative and we are looking for values that cause our expression to be less than or equal to  0 (negative or 0),  would be part of the solution.

 From the interval , I choose 1 to use  to test this interval. (I could have used 10, 25, or 10000 as long as it is in the interval)

 *Chose 1 from 3rd interval to plug in for x

 Since 2 is positive and we are looking for values that cause our quadratic expression to be less than or equal to 0 (negative or 0),  would not be part of the solution.

 Interval notation:  Graph: *Closed interval indicating all values between 1/3 and 1/2, inclusive *Visual showing all numbers between 1/3 and 1/2, inclusive

Last revised on Dec. 30, 2009 by Kim Seward.