Quadratic Inequality

[mathdad]

Solve the inequality

2x^2 - x < 10 and graph the solution set.

I know another method to solve this problem (and others like it) that has nothing to do with the textbook method for solving quadratic inequalities. However, I am always willing to learn new ideas.

Textbook Solution Steps:

1. Rearrange the inequality so that 0 is on the right side.

2x^2 - x - 10 < 0.

2. Graph f(x) = 2x^2 - x - 10 to find where f(x) is less than 0.

3. From the graph, we can see that the x-intercepts are x = 5/2 or x = -2. The y-intercept is - 10.

4. Find vertex. The vertex is
(1/4, -10.125).

The graph is below the x-axis where f(x) < 0 between x = - 2 and x = 5/2.

Since the inequality is strict, the solution is {x|-2 < x < 5/2} or, using interval notation, (-2, 5/2).

What does the author mean by "...the inequality is strict"? I prefer to factor
f(x) = 2x^2 - x - 10 and then plot the values of x on the number line. I can then select numbers from each section of the divided number line to sub into the original inequality or the factored form simply to find out which sections are true or false. From this number line information, I can then find the solution. What do you say? How do you solve quadratic inequalities? What method do you like to use?

 

[Harry_the_cat]

Solve the inequality

2x^2 - x < 10 and graph the solution set.

I know another method to solve this problem (and others like it) that has nothing to do with the textbook method for solving quadratic inequalities. However, I am always willing to learn new ideas.

Textbook Solution Steps:

1. Rearrange the inequality so that 0 is on the right side.

2x^2 - x - 10 < 0.

2. Graph f(x) = 2x^2 - x - 10 to find where f(x) is less than 0.

3. From the graph, we can see that the x-intercepts are x = 5/2 or x = -2. The y-intercept is - 10.

4. Find vertex. The vertex is
(1/4, -10.125).

The graph is below the x-axis where f(x) < 0 between x = - 2 and x = 5/2.

Since the inequality is strict, the solution is {x|-2 < x < 5/2} or, using interval notation, (-2, 5/2).

What does the author mean by "...the inequality is strict"? I prefer to factor
f(x) = 2x^2 - x - 10 and then plot the values of x on the number line. I can then select numbers from each section of the divided number line to sub into the original inequality or the factored form simply to find out which sections are true or false. From this number line information, I can then find the solution. What do you say? How do you solve quadratic inequalities? What method do you like to use?

The inequality is strict because it is \(\displaystyle <\) rather than \(\displaystyle \leq\).

 

[Otis]

... Textbook Solution Steps: ...
2. Graph f(x) = 2x^2 - x - 10 to find where f(x) is less than 0.
3. ... The y-intercept is - 10.
4. Find vertex. The vertex is (1/4, -10.125) ...

As the textbook uses a graphical method, I don't understand why their steps list the y-intercept and vertex.

Is this the same text that you mentioned earlier (that is, the book you're using for your algebra review)?

?

 

[mathdad]

As the textbook uses a graphical method, I don't understand why their steps list the y-intercept and vertex.

Is this the same text that you mentioned earlier (that is, the book you're using for your algebra review)?

?

Yes. The 9th Edition of College Algebra by Michael Sullivan. Most textbooks are not student friendly but I like Sullivan's math books. I should be using a modified college algebra textbook but Sullivan's questions are challenging and similar to classroom problems.

 

[MarkFL]

Let's pick up here:

[MATH]2x^2-x-10<0[/MATH]
I would next factor:

[MATH](2x-5)(x+2)<0[/MATH]
This gives me two critical values:

[MATH]x\in\left\{-2,\frac{5}{2}\right\}[/MATH]
As all roots of the polynomial on the LHS are of odd multiplicity, I know the sign of the polynomial will change across all critical values. So, if I know the sign of the polynomial in one interval, I know the sign in all intervals. Our two critical values divide the real number line into 3 intervals:

[MATH](-\infty,-2)[/MATH]
[MATH]\left(-2,\frac{5}{2}\right)[/MATH]
[MATH]\left(\frac{5}{2},\infty\right)[/MATH]
Let's pick a value for \(x\) from the middle interval, specifically, \(x=0\). We see by inspection that for that value of \(x\) the factored polynomial is negative, and so that middle interval is part of the solution set, and the other two intervals cannot be.

 

[Otis]

... I should be using a modified college algebra textbook ...

What kind of modification are you thinking about?

If I were solving that exercise by hand, I would use the method described by Mark in post #5 (i.e., testing values within the intervals defined by sign changes). If I were using inspection from the parabola's graph, then I wouldn't bother looking at the y-intercept or vertex because the solution would be obvious.

?

 

[mathdad]

Let's pick up here:

[MATH]2x^2-x-10<0[/MATH]
I would next factor:

[MATH](2x-5)(x+2)<0[/MATH]
This gives me two critical values:

[MATH]x\in\left\{-2,\frac{5}{2}\right\}[/MATH]
As all roots of the polynomial on the LHS are of odd multiplicity, I know the sign of the polynomial will change across all critical values. So, if I know the sign of the polynomial in one interval, I know the sign in all intervals. Our two critical values divide the real number line into 3 intervals:

[MATH](-\infty,-2)[/MATH]
[MATH]\left(-2,\frac{5}{2}\right)[/MATH]
[MATH]\left(\frac{5}{2},\infty\right)[/MATH]
Let's pick a value for \(x\) from the middle interval, specifically, \(x=0\). We see by inspection that for that value of \(x\) the factored polynomial is negative, and so that middle interval is part of the solution set, and the other two intervals cannot be.

Nicely done. I saw a video online of a tutor named Patrick. Patrick said to pretend the inequality sign is an equal sign. Just like you said, we factor the trinomial. Afterward, setting each factor to zero and solve for x. I will post the link here.

Here it is:

What do you think of the video and the method shown?

 

[mathdad]

What kind of modification are you thinking about?

This is what I mean:



If I were solving that exercise by hand, I would use the method described by Mark in post #5 (i.e., testing values within the intervals defined by sign changes). If I were using inspection from the parabola's graph, then I wouldn't bother looking at the y-intercept or vertex because the solution would be obvious.

?

 

[Harry_the_cat]

Nicely done. I saw a video online of a tutor named Patrick. Patrick said to pretend the inequality sign is an equal sign. Just like you said, we factor the trinomial. Afterward, setting each factor to zero and solve for x. I will post the link here.

Here it is:

What do you think of the video and the method shown?

Personally, I don't like this method and would never teach it. For a variety of reasons:

1. He used x=-6 as a "test" of the left region, and found that the expression was not less than 0. Then he made a gigantic leap to say that therefore all the other values less than -5 will yield a value not less than 0. That is true, but have you thought about WHY that is true? It doesn't teach "understanding"; it's just giving a recipe. He then does the same thing with x=0 in the middle region and finds x=0 yields a value less than 0, and again, without any explanation, says that all numbers in the middle region will yield values less than 0. WHY?? ( a rhetorical question!)

2. Why "test" all 3 regions? In fact, why test any regions? I agree with finding the x-intercepts. Also, since "a" is positive, the graph is a concave upwards parabola. If you quickly plot the x-intercepts, and draw a concave upwards parabola through them, it is pretty obvious that the graph will be below the x-axis (ie the expression is negative) between the two x-ints and above elsewhere. No calculations required.

 

[mathdad]

Personally, I don't like this method and would never teach it. For a variety of reasons:

1. He used x=-6 as a "test" of the left region, and found that the expression was not less than 0. Then he made a gigantic leap to say that therefore all the other values less than -5 will yield a value not less than 0. That is true, but have you thought about WHY that is true? It doesn't teach "understanding"; it's just giving a recipe. He then does the same thing with x=0 in the middle region and finds x=0 yields a value less than 0, and again, without any explanation, says that all numbers in the middle region will yield values less than 0. WHY?? ( a rhetorical question!)

2. Why "test" all 3 regions? In fact, why test any regions? I agree with finding the x-intercepts. Also, since "a" is positive, the graph is a concave upwards parabola. If you quickly plot the x-intercepts, and draw a concave upwards parabola through them, it is pretty obvious that the graph will be below the x-axis (ie the expression is negative) between the two x-ints and above elsewhere. No calculations required.

Patrick has a math degree. Patrick does not teach the theory, the why in most of his video lessons. He basically helps students with what they mainly seek--- how to solve problems. I prefer watching professor Leonard. Leonard not only explains how to solve problems. He also gives THE WHY in all his college lectures. You can find him on youtube.com.

 

[Otis]

Here's another way to think about the inequality: consider the sign of each factor.

(x + 2)(2x - 5) < 0

For the left-hand side to be negative, the signs of the factors must be opposite. That gives two cases: Either x+2 is negative and 2x-5 is positive, or it's the other way around. The first case

x + 2 < 0 AND 2x - 5 > 0 leads to x < -2 AND x > 5/2

But that's impossible. No number can be both negative AND positive, so the second case is true.

x + 2 > 0 AND 2x - 5 < 0 leads to x > -2 AND x < 5/2

?

 

[Harry_the_cat]

Patrick has a math degree.

But does he have an education degree?

Patrick does not teach the theory, the why in most of his video lessons. He basically helps students with what they mainly seek--- how to solve problems.

I disagree here. He is not teaching how to solve problems - he is teaching the steps to solve a particular problem. Without any understanding, the student cannot apply that to other problems which are slightly different.

I prefer watching professor Leonard. Leonard not only explains how to solve problems. He also gives THE WHY in all his college lectures.

That's great. The WHY is so important.

You can find him on youtube.com.

 

[mathdad]

Here's another way to think about the inequality: consider the sign of each factor.

(x + 2)(2x - 5) < 0

For the left-hand side to be negative, the signs of the factors must be opposite. That gives two cases: Either x+2 is negative and 2x-5 is positive, or it's the other way around. The first case

x + 2 < 0 AND 2x - 5 > 0 leads to x < -2 AND x > 5/2

But that's impossible. No number can be both negative AND positive, so the second case is true.

x + 2 > 0 AND 2x - 5 < 0 leads to x > -2 AND x < 5/2

?

Interesting notes.

 

[mathdad]

Patrick is a tutor. He has lots of videos including calculus 1-3 and linear algebra but I prefer professor Leonard. BTW, I also like math tutor Jason Gibson. Both excellent at teaching math. Of course, professor Leonard is an actual college professor in California.

He is currently uploading an entire differential equations course one chapter at a time. He then plans to do the same with linear algebra, trigonometry, geometry, college algebra, precalculus, etc. He looks like Clark Kent. Look him up on youtube.

 

[JeffM]

Nicely done. I saw a video online of a tutor named Patrick. Patrick said to pretend the inequality sign is an equal sign. Just like you said, we factor the trinomial. Afterward, setting each factor to zero and solve for x. I will post the link here.

Here it is:

What do you think of the video and the method shown?

Patrick's method is exactly what I teach those whom I tutor although I explain it a bit differently. The method works for any continuous function.

You are asked to find the interval where f(x) R 0. R here stands for less than, less than or equal, greater than, or greater than or equal. Although this appears to be limited, it is not. Finding the interval where
g(x) R h(x) is the same problem LOGICALLY as finding the interval where
f(x) R 0 given f(x) = g(x) - f(x).

Step 1: Determine the values where f(x) = 0. I usually say that as solve the related equality.

Step 2: If R is less than or equal or greater than or equal, the values of equality found are INCLUDED endpoints. If R is less than or greater than, the values of equality found are EXCLUDED endpoints.

Step 3: Test a value less than the smallest endpoint, a value greater than the largest endpoint, and a value between each of the endpoints. For example, if you find four points of equality, you must test five points.

Step 4: Use the results of the test to determine which intervals satisfy R.

 

[mathdad]

Patrick's method is exactly what I teach those whom I tutor although I explain it a bit differently. The method works for any continuous function.

You are asked to find the interval where f(x) R 0. R here stands for less than, less than or equal, greater than, or greater than or equal. Although this appears to be limited, it is not. Finding the interval where
g(x) R h(x) is the same problem LOGICALLY as finding the interval where
f(x) R 0 given f(x) = g(x) - f(x).

Step 1: Determine the values where f(x) = 0. I usually say that as solve the related equality.

Step 2: If R is less than or equal or greater than or equal, the values of equality found are INCLUDED endpoints. If R is less than or greater than, the values of equality found are EXCLUDED endpoints.

Step 3: Test a value less than the smallest endpoint, a value greater than the largest endpoint, and a value between each of the endpoints. For example, if you find four points of equality, you must test five points.

Step 4: Use the results of the test to determine which intervals satisfy R.

1. What do you mean by CONTINUOUS FUNCTION? Are you saying ""continuous" as taught in calculus 1?

2. Can you apply your steps to the following problem below?

Solve the inequality 2x^2 + 12x + 6 > 0.

 

[JeffM]

Yes, I mean continuous as it is defined in standard mathematics.

By the quadratic formula, the related equality has solutions at x = -3 + sqrt(6) and -3 - sqrt(6).

So those two points are excluded.

Testing a point below -3 - sqrt(6), I choose - 10. So f(-10) = 200 - 120 + 6 > 0.

Testing a point above -3 + sqrt(6), I choose 10. So f(10) = 200 + 120 + 6 > 0.

Testing a point between them, I choose - 3. So f(-3) = 18 - 36 + 6 = -12 < 0.

f(x) > 0 if x < -3 - sqrt(6) or x > -3 + sqrt(6).

 

[mathdad]

Yes, I mean continuous as it is defined in standard mathematics.

By the quadratic formula, the related equality has solutions at x = -3 + sqrt(6) and -3 - sqrt(6).

So those two points are excluded.

Testing a point below -3 - sqrt(6), I choose - 10. So f(-10) = 200 - 120 + 6 > 0.

Testing a point above -3 + sqrt(6), I choose 10. So f(10) = 200 + 120 + 6 > 0.

Testing a point between them, I choose - 3. So f(-3) = 18 - 36 + 6 = -12 < 0.

f(x) > 0 if x < -3 - sqrt(6) or x > -3 + sqrt(6).

Good job.

 

[Steven G]

Patrick has a math degree. Patrick does not teach the theory, the why in most of his video lessons. He basically helps students with what they mainly seek--- how to solve problems. I prefer watching professor Leonard. Leonard not only explains how to solve problems. He also gives THE WHY in all his college lectures. You can find him on youtube.com.

Use these videos to study from: https://www.youtube.com/user/FreeAcademyForMath

 

[Steven G]

I hate testing values and never like using it.
I ask my class when is x+5 >0? Students can do this and say when x>-5. So in the standard table that everyone uses here they put + for the intervals where x>-5 and negative signs elsewhere.
I then ask when x-4>0. Students respond when x>4. They then put + signs in all intervals when x>4 and - signs elsewhere.

In this way you do not have to test every cp in each and every interval.

And yes, for the students who can't solve these linear inequalities in their head I do algebraically solve things like x+5>0 and x-4>0 for x.