Solve each quadratic inequality. Give the solution set in interva...

Question

Solve each quadratic inequality. Give the solution set in interval notation. See Exam-ples 5 and 6. x^2-2x≤1

Accepted Answer

Hello, everyone. In this video, we're going to be looking at this practice problem which states for the following quadratic inequality solve and express the solution set in interval notation. And the quadratic inequality is X squared minus six X less than or equal to six. And for this problem, we're given four answer choices A B C and D and you can see that A and B both have brackets, we'll see has parentheses and then D is no solution. So when we look at our inequality, we have back squared minus six X less than or equal to six. And you can see that we have less than or equal to, which is usually associated with brackets in interval notation and not parentheses. So it seems like A or B would be the correct intervals. But let's go ahead and solve the solution set to see if that is true. So I'm going to start by moving all my terms to the left hand side. So also tracked by six and that leaves me with X squared minus six X minus six is less than or equal to zero. And now you can see I have this equation that looks very similar to a quadratic equation. So in order to solve for X, I'm gonna try completing the square, which means I'm going to be grouping X squared and minus six X together. So when I do that, it's going to become X squared minus six X plus. Remember that when I want to complete the square, I'm looking to add an integer in here so that I end up with a factor like this. So that's what we're looking to do by completing the square. But I still have the -6 less than or equal to zero on the outside. And the number that I need to add here is often denoted A squared. And to solve for A, I need this middle coefficient here. So to A is equal to the middle coefficient six. So if I solve for A, I get that A is equal to three, so that means that the number I need to add is three squared or nine. And so that means that my factor down here, the first term would be X plus nine. We don't need nine here, but we need is actually a and we already saw for a, we know that A is three, but going back up here, we added a nine inside the parentheses and we can't just add numbers into equations. So because I added a nine inside the parentheses, I need to subtract a nine outside the parentheses. So this will become -6 -9, less than or equal to zero. So if I carry this portion Down here, now I have -6 -9 is -15, less than or equal to zero. And now you can see that we have X without any power. So I'm going to start to solve for X to find the solution set. So go ahead and move it up here. So I'll have X plus three squared minus 15, less than or equal to zero. And in order to start solving for X, I'm going to add 15 to both sides. And now I have X plus three squared less than or equal to 15. So in order to get rid of the squared on the left hand side, I'm going to take the square root of both sides. And in this case, because I was taken to the power of two, which is an even power I need to rewrite this as negative square root of 15, less than or equal to X plus three. Notice we got rid of the squared because we took the square root And less than or equal to positive 15. And actually a slight modification here is that this factor X plus three actually needs to be X minus three because the middle term here, The -6 is negative. So that means that this portion here should be minus. You can also check this by doing the foil method. We have X minus three. Times X minus three. So the first terms multiplied together X squared, the outer terms X times negative three is negative three, X negative three times X is negative three, X and negative three times negative three is positive nine. So you can see that if I combine these middle terms, I have negative six x so I need to change my factor to minus here as well as here. And then finally, also here. So now you can see that in order to isolate X, I need to add three. So if I add three to all these inequalities, I have three minus the square root of 15, less than or equal to X, which is less than or equal to three plus the square root of 15. And from here, you can see that the solutions for X or X is sandwiched between two boundaries, we have this lower boundary and this upper boundary. So if I were to write this in interval notation, we also have the less than or equal to in both of these. So because I have the equal to, I know it's going to be a bracket. And my lower boundary is three minus the square root of 15, All the way to three plus the square root of positive again, and then another bracket. So this interval becomes my solution set for this inequality. And if we look at the answer choices, you can see that answer choice A also has the same interval of rocket three minus the square root of 15 to 3 plus the square root of 15 and, and bracket. So hopefully this video helped and see you next time.

Solve each quadratic inequality. Give the solution set in interval notation. See Exam-ples 5 and 6. x^2-2x≤1

Key Concepts

Quadratic Inequalities

Interval Notation

Test Points

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