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+4x>-1

Accepted Answer

Hello everyone. In this video, we're going to be looking at this practice problem that states for the following quadratic inequality solve and express the solution set in interval notation. And the quadratic inequality is X squared plus six, X is greater than negative six. And were given four answer choices for the solution set answer choice A is negative three minus the square root of 32, negative three plus the square root of three answer. Choice B is negative infinity to negative three minus the square root of three union with negative three plus square root of three to positive infinity. Answer choice C is negative infinity to positive infinity which is all real numbers. And answer choice D is no solution. So in this case, let's take a look at our quadratic inequality X squared plus six, X is greater than negative six. In order to find the solution set, we need to solve for X in this inequality. So because we're dealing with a quadratic inequality, we can solve for this X value as A as we would solve for X and, and quadratic equation. So I'm going to move the six over from the right hand side over to the left hand side by adding six to both sides. So this becomes X squared plus six, X plus six is greater than zero. And now you can see I have a quadratic equation and I'm going to try to solve for X by completing the square. So in order to complete the square, I'm gonna want a group X squared and six X together. So I'm going to have X squared plus six X plus. This is the blank where I need to complete the square, but I still have the plus six on the outside of that grouping and I still have the greater than zero portion. So now to find this blink, I need to find numbers that multiply or add to six. So I need the numbers which I call a to a, needs to Equal six. So if I solve for a, in this course, in this case, by dividing by two, I get a is equal to three. Well, this blank that I need to solve for is a squared and we just solved for a. So it's three squared and three squared is nine. But because I added this nine inside the parentheses, I just can't, I can't just add a value into an equation. So I need to make sure to subtract it on the outside here. So this becomes parentheses X squared plus six X plus nine and parentheses plus six minus nine. So now I completed the square and I can factor my quadratic equation. So when I factor, I want the first terms of my factors to multiply to get X squared. So X times X is X squared. And I want my second terms to add to six, but multiply two positive nine. So I can do X plus three. And now if I do a full multiplication, I have X times three is three, X and then I'm going to have three times X is three X and then finally, three times three is nine. If I combine these middle terms, three, X plus three, X is six X, and you can see that gives me back fee equation in the parentheses. So I have factored correctly and notice that X plus three is the same as X plus three. So I can write this factor as X plus three squared. And then keep in mind I still have plus 6 -9 is negative three, greater than zero. So now I can finally start to isolate X. So I'm going to start by adding three to both sides. So this becomes X plus three squared is greater than three. If I take the square root of both sides, The left hand side becomes X-plus three. And my right hand side is now greater than plus or minus square root of three. So in this case, you can see that we have to sort of situations coming out. I could have X plus three is greater than positive square root of three or I could have X plus three is greater than negative square root of three. And if I move negative over by dividing by negative one, this inequality becomes negative X plus three. And I have to switch the inequality because I divided by a negative across. So negative X plus three is now less than the square root of three. So now I can continue to solve for X in these two variations. So on the left hand side or this first inequality, I'll subtract three from both sides. So that leaves me with X is greater than a negative three plus the square root of three. And that is X isolated. So that is one of my solutions. And for this second inequality, I can distribute the negative. So this becomes negative X -3 is less than the square root of three. And once again, I'll add, I'll move the three over this case, I need to add three. So that leaves me with negative X is less than three plus the square root of three, but I still have this negative. So once again, I'm going to move it across the inequality by dividing by negative one on both sides. So that leaves me with X is less than three divided by negative one is negative three and positive square root of three divided by negative one is negative square root of three. So now I solved my second inequality. And now I can start to combine these two to find my solution set for X. So I'm gonna draw a number line and I'll label this point zero and this point negative three minus square root of three, actually, I'm going to move the zero over to the right. And this line or this tick here will be negative three plus the square root of three. So if I look at my first interval here where I have X is greater than negative three plus the square root of three, notice how it's greater than and not greater than or equal to. So it's going to be an open circle. And because X is greater, this interval is going to go to the right of the point. So it's going from negative three plus the square root of three to positive infinity. And if I look at my second interval where I have X is less than negative three minus the square root of three, again, I have an open circle but now X is less than that number. So I need to go to the left. So that means that portion is going to negative infinity. So now that I have it on this number line, I can more easily convert it into interval notation where X goes from negative infinity to negative three minus the square root of three. And then there's this gap here. But there's another interval from negative three plus the square root of three to positive infinity. And I'll write that more clearly up here and notice how all of these are parentheses because we only have greater than, or less than, and not greater than, or equal to, or less than, or equal to. So this interval here becomes my final answer for the solution set for this problem expressed an interval notation. And if we look at the answer choices, you can see that answer choice B also has the interval listed as negative infinity to negative three minus the square root of three union with negative three plus the square root of positive three to positive infinity. So hopefully this video hoped and see you next time.

Solve each quadratic inequality. Give the solution set in interval notation. See Exam-ples 5 and 6. x^2+4x>-1

Key Concepts

Quadratic Inequalities

Interval Notation

Graphing Quadratics

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