in-12-17-use-a-graph-to-find-the-solution-set-of-each-inequality-x2-2-x1-0

Question

In $12-17,$ use a graph to find the solution set of each inequality.
$$
x^{2}-2 x+1 < 0
$$

Finian Lickona · Accepted Answer

All right. So for this exercise, we are given the inequality X squared minus X minus two is greater than zero. And we are asked to use a graft to find the solution. Set this inequality. So I have the graft generated already right here and now what we need to do is observe the graph and determine which X values create a functional output that is greater than zero. So if we were to plug in zero into our function, we would get an output of negative too. So the zero clearly doesn&#x27;t work. Likewise with any of the values on the interior of this problem. However, all of the values outside of the parabola in either direction very clearly generate functional outputs that are greater than zero. So our solution set is going to be all the points in this region and this region. Okay?

Finian Lickona · Answer

Okay, So in this exercise, we are given the any equality. X squared minus two X plus three. Oh, no, I&#x27;m sorry. It should be. Plus two X minus three is less than zero. And I&#x27;ve already generated the graph of this particular polynomial right here. But just for clarity. We&#x27;re gonna go ahead and factor anyways. So this factors into X minus one X plus three. And if we were to expand this, we get X squared minus a plus three x minus X minus three Anyways. Okay. So what we are looking for is the region of X values and their corresponding. Why values that satisfy this inequality? So if I were to pick, let&#x27;s say, um the number negative three, and plug this into this equation. I get negative. Three squared, plus two times negative. Three minus three equals nine minus six. Minus three equals zero. And technically, zero is not strictly less than zero. So I can&#x27;t include the value Negative three, but I can include any value. Ah, after that. So I&#x27;m gonna put an open dot right there to indicate a open interval. Now, if I were to pick any of these remaining X values like, let&#x27;s say negative to. For example, I do negative two squared plus two times, thank you. Two minus three. He goes for minus four. Minus three equals negative three, which is less than zero. So that means that this value *** too, is going to satisfy our inequality. And I hope it&#x27;s pretty clear that all of the values from negative three up to positive one are going to satisfy the inequality given here because all their outputs, like all their functional outputs, are below the X axis axis, which means they&#x27;re less than zero. So our solution region is actually going to be this entire portion of the proble below the X axis, okay?

Finian Lickona · Answer

Okay, So for this exercise, we are given the inning quality negative. X squared plus six X minus five is less than zero. And I already have the graph of this&#x27;ll generated over here already. So we need to determine which X values create functional outputs that put the parade below below zero. So if we look at, like, say the value to we see that this output generates like, uh, looks like it&#x27;s about three, that&#x27;s a positive value that&#x27;s greater than zero, not less than zero. So we need to be doing is looking at all of the X values that create outputs below zero. That&#x27;s going to be all the values from it Looks like that&#x27;s a positive one, and less than that. And then here we also have a positive five and greater than that. So our solution areas are actually gonna be everything over here. Hand everything over here. Okay?

Sherrie Fenner · Answer

question 27. I have X players by its X, my guest Well, over X minus one. It&#x27;s less than zero. So I&#x27;m gonna find my critical numbers. My first critical number. I know it&#x27;s gonna be one because my denominator cannot equal zero. So one minus one in my denominator, So critical number one set my denominator equal to zero. So expert e one as my first critical number then my numerator I like instructor it insanity folded, See grow. So I have X and X equals X squared. And now I&#x27;m saying, OK, the signs air different because my last sign is a negative story of a minus and a plus. The larger number goes with the minus. So what? Most places give me 12 and some tracks. Give me 14 and three and the larger number goes with the negative sign. So now I have X plus three equals zero or X minus, or equal zero. Subtract three, subtract three. So X equals negative free at or at or or X equals four. So my critical numbers Air one negative. Three and four one four Negative three. So I have to pick a point to the left of negative three. So my pick negative four trial numbers I&#x27;ll try and negative four. I&#x27;ll try in between three negative three and one zero in between one and four. Let&#x27;s go to and to the right of four with five. So now four split your negative or weird minus negative for minus well over. Negative for minus one. Is that less than zero 16 plus or minus 12 16 plus who were minus Well, so 16 plus fours 20 minus 12 is eight over negative five. But that&#x27;s a negative number. That would be lessons. So that is true. So in between three and one, let me put zero in so zero. So I have negative 12 over negative one. That negative of a negative is positive. So this part here would be false. I&#x27;m gonna try to squared is for minus two minus Well, over to minus one, Which is one is that lesson zero? So that would be negative. So that is true. And to the right of four plugging by. We have 25 minus five, minus 12 over five minus one. So is that negative? That lesson zero 25 Note. That&#x27;s gonna be positive. So this is gonna be 28 over four, which is a positive number. Falls. So my regions here I have a friend didn&#x27;t shade it this way because all those numbers would be true. At one. I would have a friend because that&#x27;s where equals zero, for I have a friend because it&#x27;s less than in that equal missy row and I shave in between. So my, uh, interval notation would be from negative infinity, the negative three friend you get in one before.

Peter Winans · Answer

All right, so we have X squared. Plus two X is less than zero, and we want to find the solution set that makes this sentence true. Well, we have to solve this. Like for solving a quadratic function are a quadratic equation that is and notice. There&#x27;s only two terms here, so I can fact around a common factor of X, and that will leave me with X Times X plus two less than zero. So if I take each factor is my next step and let it equals zero, I get X equals zero and I get X Plus two zero. So we get X equals negative too. So we have our two routes now. Usually what we want to do is pick of value. That&#x27;s between these and determine what our solution said stealing. In other words, if I pick a value between zero and negative too, and it tax than I know, my solution set is all the numbers between zero and negative too. So we&#x27;re gonna test X equals negative one and see what happens. I&#x27;m gonna have negative one squared plus two times negative one. And if this is in the solution said then I&#x27;ll get an answer that&#x27;s less than zero. So that&#x27;s gonna be one minus two, which is negative. One negative one is definitely less than zero. So that indicates to me that all the values between zero and negative to not including this year on the negative, too, because notice we only want to be less than zero right now. So my solution sets going to be. The values between zero and two are, in other words, the interval from 0 to 2. Excuse me, that&#x27;s a negative too. So if we&#x27;re between zero and negative too, that&#x27;s gonna be the interval from negative to 20 So I have a negative two. We have a zero on our number line and our solution sat will be everything in between negative two and zero.

Problem

In $12-17,$ use a graph to find the solution set …

Problem 15 Easy Difficulty

In $12-17,$ use a graph to find the solution set of each inequality.
$$
x^{2}-2 x+1 < 0
$$

Answer

There is no real number solution to the given inequality.

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Ann Xavier Gantert

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Absolute Value - Example 1

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There is no real number solution to the given inequality.

Algebra 2 and Trigonometry

Catherine R.

Alayna H.

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Absolute Value - Example 1

Absolute Value - Example 2

Ann Xavier GantertAlgebra 2 and Trigonometry

Catherine R.

Alayna H.

Kristen K.

Michael J.

Ann Xavier Gantert

Algebra 2 and Trigonometry