Use the technique described in Exercises 87–90 to solve each ineq...

Question

Use the technique described in Exercises 87–90 to solve each inequality. Write the solution set in interval notation. -x^2 + 4x + 1 ≥ 0

Accepted Answer

Hello everyone. In this video, we're going to be looking at this practice problem which states find the solution for the given quadratic inequality express the answer in interval notation. And the inequality is negative X squared plus six X plus one is greater than or equal to zero. And were given four answer choices. A is an interval from three minus the square root of 10 to three plus the square root of 10. B is another interval interval from negative seven to positive 13 C is another interval from negative three minus the square root of 10 to negative three plus the square root of 10 and D is another interval from three minus the square root of 10 to 3 plus the square root of 10. And you can see that A and D differ not because of their boundaries, but because D has parentheses instead of brackets. So in order to figure out which of these is correct, let's look at our inequality which is negative X squared plus six X plus one greater than or equal to zero. So to find our solution, we need to isolate X and to start doing that, I'm going to get rid of this negative in front of the X squared by dividing both sides by negative one. So if we divide both sides by negative one and left with positive X squared minus six X minus one greater than or equal to zero, divided by negative one is still zero. And I'm going to solve for X by first completing the square. So in order to complete the square, I'm going to group the X squared and the negative six X together. So I'm going to have X squared -6 X plus the number I'm going to complete the square with and I still have the negative one and greater than or equal to zero. So in order to complete the square, the number that I need to complete the square with is A squared. Well, what is a, a depends on this middle coefficient here. And in this case, the coefficient is six and that is equal to two a. So if I sell for a, by dividing by two, I get A is equal to three, Which means that the number I'm going to be adding is three squared, which is nine. But because I'm adding nine into this equation, I can't just add any number into an equation. I need to make sure that I take it out Here. So I added nine inside which means I need to subtract nine on the outside of this parenthesis. So my equation becomes X squared minus six X plus nine. And now I have negative one minus nine, which is negative 10, greater than or equal to zero. And the reason that I completed the square was because I wanted to factor my quadratic equation. So when I factor this quadratic equation, I'll have X minus A squared and this is a minus because again, the middle coefficient is negative. So because it's negative, I need to make sure that it's Minour between my two terms. So that means that when I'm factor my equation, I'll have X minus A which is three squared -10 greater than or equal to zero. And from here, I can start solving for X. So I'm gonna add 10 to both sides. So that leaves me with, we'll move it up here to the right, That leaves me with X -3 squared Greater than or equal to 10. And a small adjustment here is noticed that we divided by a negative across the inequality in this first step, which means that my inequality should have changed directions instead of greater than it should be less than. So if I change these to less, then My equation is X - squared is less than or equal to 10. And now to continue isolating X, I need to take the square root of both sides. And when I do that, I have X -3 is less than or equal to the square root of 10. But because I took the square root X -3 is also greater than Negative Square root of 10. So now when I sell for X, I need to add three to both sides of this inequality. So I'm left with three minus the square root of 10 Is less than or equal to X which is less than or equal to 3-plus the square root of 10. So now you can see I have this lower boundary and this upper boundary. So if I write that in interval notation, notice how both of these and points will be, will have brackets because they're both less than or equal to and because of the equal to portion, I know that I can do a bracket. So my lower boundary is three minus the square root of Up to my upper boundary which is 3-plus the square root of 10. And again another bracket on that end. So this becomes my final solution for this problem. And you can see that answer choice. A also has the same interval from three minus square root of 10 to 3 plus the square root of 10. So hopefully this video helped and see you next time.

Use the technique described in Exercises 87–90 to solve each inequality. Write the solution set in interval notation. -x^2 + 4x + 1 ≥ 0

Key Concepts

Inequalities

Quadratic Functions

Interval Notation

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