write-the-solution-set-of-each-inequality-if-x-is-an-element-of-the-set-of-integers-x2-7-x100

Question

Write the solution set of each inequality if x is an element of the set of integers.
$x^{2}-7 x+10>0$

M S · Accepted Answer

So we have X squared minus seven X plus two. He&#x27;s got it in his ear. So 1st 1 back to our left hand side. So you&#x27;re gonna want to part our coefficient times there last time when we get positive soon and then our middle term, which we want them to add to his native seven. So now a list of factors of 10 1 intend in two and five. And because it&#x27;s positive, they both either beat need to be positive or they both need to be negative. But because our seven it&#x27;s negative, you know, they both need to be negative. So in this case, we&#x27;re choosing negative to a negative. But because I asked the negative seven so, uh, split the middle turn. So X squared minus two x minus five picks been plus tune and they were good. The 1st 2 and group the second to know we factor out the greatest common factor. So bloody X, we have X minus two and pull out a negative five. Here we have X minus two on the inside. And we know that because this is the same as this we have are two factors so excellent is to an expert in a spot of is greater than zero. So because this isn&#x27;t any quality, it takes an extra step to solve for X because you have to find a range of X values and because a positive times a positive gives us a positive and we want we want our back there&#x27;s to be multiply and give us a positive number. But because a negative times negative also gives us a positive number we need to solve. For Wayne, the factors are less than zero, so I also need to saw or X minus two times X minus pi. It&#x27;s less than zero because when I vote, if you multiply them both and it gives you a negative number, it turns to a positive because negative times negative give us a positive so we&#x27;ll have X minus two is greater than zero over here, and X minus five is greater than zero. And we acted about. Sides here still have excess grated into, and we add five to both sides here, so we&#x27;ll have exits guarded in pots. So okay, he&#x27;s a dreidel, A little number line. To combine. These two qualifications will have to hear and it&#x27;s greater dance. It goes this way, and we have five here and it&#x27;s great. And so it goes this way and the only one the only way to combine this is it. X is greater than by. So now we need to software or the wind still have X minus two is less than zero an X minus. Pi is less than zero and we add to the both sides. So accept less than two. And we had Bob to both sides and exits list in five. So I gave the door a little number line. We&#x27;ll have to and five fixes listen to So it goes this way and excess lesson five. So it goes this way and it means both qualifications when exist, list into. So that&#x27;s how we come by on that one. So because our shading goes to different directions we&#x27;re gonna use or because it&#x27;s not in between so we&#x27;re gonna use or so X is gonna get by or X is less than two. We also want to find a solution set. So I just do two numbers and I have been below so huh ex congee three. No, not through work it has to be greater than five so two numbers guarded by this 67 and then two numbers below to ISS one and zero, so that could be a solution said.

Bahar Tehranipoor · Answer

told this inequality and find the solution set. So here we want to dio is factor extra plus X minus two. So let&#x27;s do that. If we do that, we have these two terms of X. So if we have negative to really, the only thing we can factor that into is to win one. So let&#x27;s put it positive, too, and a negative one, because then we&#x27;ll have X squared plus two x minus X, which gives us this positive eggs minus two. This Quimby listen zero. So let&#x27;s find the values that make this expression zero that would be at X equals one and X equals negative, too. Now we&#x27;re going to use a number line. So our values were one and negative. To put one, I need to know to find the solution set. The easiest thing we could do is take values over one between negative two and one and then the load eight of two. So let&#x27;s start with over. One will pick something like two. We plug in to we get to minus one, which is a positive number, and then two plus two is a part of number, and that is not less than zero. So we will call back and say that this region does not work. Now let&#x27;s pick a number between eight of two and one like zero butt. Plug that and we get zero minus one, which is negative number negative 10 post to posit number two. Those numbers will be together or negative, which is less than zero. So that does in fact work. So this region is right. No, for pick something below native to like negative three. We get a negative number here and a negative number here. Multi together we get positive and that&#x27;s not less than zero. So this region does not work. So all in all our solution set is that X is going to be greater than negative too, but less than one. And that is our set.

Bahar Tehranipoor · Answer

in this exercise, we want to bite the solution. Sentence inequality. So let&#x27;s factor extra minus two x Easy thing to do to take out a next. So we have X Times X minus two. Screaming Eagle zero. Now let&#x27;s find the values of except Make this inequality zero and that will be at zero and two. Now let&#x27;s use our trusty number line. So zero to and I was just testing values that are greater than two between your own too, and in less than zero. So if we pick a number over to like three, what will we get? We&#x27;ll get a positive number three times three minus two, which is positive, which is in fact, greater than zero. So we can say that yes, this region works. Let&#x27;s pick one in betweens your own to like. One will do. One times one minus two, which is a negative numbers to this region does not work. And then we can pick a value less than zero like negative one. Plug that and we get a negative times negative, which is in fact, Raylan zero. So you can say this region works, so we also know that zero into themselves will work because we&#x27;re saying greater than or equal to zero. So we can say that X that is less than or equal to zero will give us the solution and then X, that is greater than or equal to two will give us a solution, and that is all there is to it.

Bahar Tehranipoor · Answer

in this exercise, we want to write The solution set of each to quality X is an element of the set of images. So we have X squared plus five X plus six is less than zero. So it&#x27;s factor this knowing we have a constant six year means. We have to look at the factors of six so free and two of six and one, and here we can use three into because we&#x27;ll have extra story. Times X plus two is less than zero. So are critical values of X equals negative three and negative too. So let&#x27;s draw our number line. You have negative three. We have negative too. Let&#x27;s pick values in each of these three regions, so pick something above negative too. We could just be like zero. We plug in zero, we will not get a negative numbers. This region very clearly does not work. Now we pick something between a three legged too, like negative 2.5 will get a positive times negative, which is less than zero. So this region does work. Now we pick something below negative three, like negative four will get a negative negative, which is positive. So this region does not work. Now we have exes between negative three and negative, too. But because this question is asking for access and element on the set of managers, you can see there&#x27;s actually no images that go between any 23 and go to because they&#x27;re not inclusive. We can&#x27;t use negative three negative, too, because plugging those and we&#x27;ll get zero. But we&#x27;re just looking for less than zero, so there&#x27;s actually no injures in this solution. Set satisfied inequality.

Angelina Chavez · Answer

in this problem were asked to solve the quadratic inequality and write the solution set in interval notation, we can start solving our inequality by factoring the quadratic that we have on the left hand side. So when we factor this, we kill it. The quantity X minus five times the quantity X minus two not is greater than zero. We can then set this equation on the left hand side, equal to zero to determine our endpoints. So when we do this, we just have everything on our left hand side is equal to zero and we can see that this is true when X is equal to positive two and positive five. So now we can set up a table where we can break up a number line into intervals to determine what our solutions that is. So here we have our table and we can note that are intervals are dependent on the end points that we determined by factoring or inequality. We can use test points that air within each interval, but not equal to our endpoints to determine and values within these intervals make our inequality true, or Fels. And if the inequality is true, within an interval. That means that this interval is part of our solution set. So we can start with our first interval here from negative infinity to two. And we have a test point of zero and we&#x27;ll go ahead and plug that into our inequality shown here. So we have negative five times Negative two and this is greater than zero. So we end up getting that positive. 10 is greater than zero. And this is true. We can repeat the same procedure for our second test point and second interval. So we get negative one times positive. Two is greater than zero. We get negative, two is greater than zero and this is false. And then we can just repeat the same procedure for our last test point. And then we will determine out of all of the true and false answers which are in our solution set. So we see that out of our three intervals, these two intervals here are true. Which means that they&#x27;re part of our solution set. This is not part of our solution set this interval here because we saw that this made rne quality false. So our solution set mrs follows and we use soft brackets because we do not have an equal to component in our original inequality. So here we have our final solution set.

Mitchell Cutler · Answer

All right. Welcome back. This bone, we have X squared. Um, minus seven X. Kristen. It&#x27;s a sanity. Coaches zero call this problem. We first need Teoh factor the left hand side here this week, in fact, that as X when it&#x27;s five times X minus two. Next we can find what I use. A X would make this equation equal to zero, which in this case is X is equal to five. Next is equal. That you So these are the roots of the quadratic equation. Um, we get that this from X if X is he qualified in this term right here. Become zero, of course. Is, uh, to this dream becomes zero. All right, Now, if we garden number line to here five here And since there are no repeated roots in this quadratic equation, um, you know that, uh, it&#x27;s gonna automate between positive and negative, uh, be from before to between two. And five and after five. So if we plug in, let&#x27;s say zero for the situation, which is over here. I agree. Your service. Uh, let&#x27;s say it makes this equation positive. You know that this would be negative, and this would be positive. If, on the other hand, it is negative. Um, and this would be positive. This would be negative. All right, What&#x27;s my know what happens? All right, So we plug in Joke for X. Does your minds five zero ministry that is equal to negative. Five times negative, too. Which is equal to 10 which is greater than zero. This is positive. Negative. Positive groups. Positive. Right there. That means that, uh, this equation are the The quality is going to be true. Um, on the interval, who should be a square bracket? Because it includes two from 2 to 5.

Problem

Write the solution set of each inequality if x is…

Problem 6 Medium Difficulty

Write the solution set of each inequality if x is an element of the set of integers.
$x^{2}-7 x+10>0$

Answer

$\{x | x<2 \text { or } x>5\},\{\ldots ., 1,2,6,7, \ldots\}$

Related Courses

Algebra 2 and Trigonometry

Related Topics

Discussion

Top Algebra Educators

Alayna H.

Caleb E.

Kristen K.

Michael J.

Algebra Courses

Absolute Value - Example 1

Absolute Value - Example 2

Recommended Videos

Watch More Solved Questions in Chapter 1

Video Transcript

Ann Xavier Gantert

Algebra 2 and Trigonometry

Alayna H.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\{x | x<2 \text { or } x>5\},\{\ldots ., 1,2,6,7, \ldots\}$

Algebra 2 and Trigonometry

Alayna H.

Caleb E.

Kristen K.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

Ann Xavier GantertAlgebra 2 and Trigonometry

Alayna H.

Caleb E.

Kristen K.

Michael J.

Ann Xavier Gantert

Algebra 2 and Trigonometry