Solve each inequality. Give the solution set using interval notat...

Question

Solve each inequality. Give the solution set using interval notation. x²-3x ≥ 5

Accepted Answer

Hey, everyone here, we are asked to find the solution set of the following inequality and express it in interval notation. And we are given the inequality X squared minus five X is greater than or equal to eight. Now, our first step here in finding the solution set is to make the right hand side of our inequality just zero. So we just need to sit attract eight from both sides. And now rewriting our inequality, we have X squared minus five X minus eight is greater than or equal to zero. Now, our next step is to find values for X by setting the left side of our inequality equal to zero. So doing this, we have X squared minus five, X minus eight is equal to zero. And now looking at our equation, we see that we have a quadratic equation. So to find values for X, we need to recall the quadratic formula which tells us that X is equal to negative B plus or minus the square root of B squared minus four A C all over two times A. So now we just need to identify our values for A B and C according to our equation. So first for the value for a, well, A is just the coefficient of X squared. So in our case, A is just one now for the value of B will be, it's just the coefficient of X. So B is negative five and now C is the value of the constant in our equation which is negative eight. So C is negative eight. Now with A B and C identified, we can go ahead and just substitute in their values into the quadratic formula to find our values for X. So here we will have X is equal to the negative quantity of negative five plus or minus the square root of the quantity of negative five squared minus four times one times negative eight. And this is all divided by two times one. And now simplifying here, we will have positive five plus or minus the square root of 57 all divided by two. And now from the quadratic formula, we see that we have come up with two solutions for X, one with a positive square root and one with a negative square root. So with this, we can write out our two solutions for X. So our first solution, we can choose the positive square root. So we have the positive square root of 57 plus five all divided by two. And then our second solution is the negative square root of 57 plus five all divided by two. And now, with these two values for X that we found, we can go ahead and write them as intervals to show that the X axis can be separated into the intervals where we go first from negative infinity, then to the smallest solution that we found, which is the negative square root of 57 plus five all divided by two. And both of these values for now are closed by parentheses. And now we go from our solution of negative square root of 57 plus five all divided by two. We go from this value to the positive square root of 57 plus five all divided by two. And lastly, we go from the largest solution value of X where we had the positive square root of 57 plus five all divided by two. We go from this value all the way to positive infinity. And again, just like the other intervals, these values are closed by. So now the next step is to test each of these intervals to see which of them serve as a solution to our inequality. So our first step is to let F of X be equal to the left side of our inequality where we had X squared minus five, X minus eight. And now we want to set up a table where we first take an interval, one of the intervals that we identified. And then we select a test value that fits within the interval. And with this test value, we then substitute it in two F of X. And from the value we obtain from F of X, we can form a conclusion based off of its sign. So now we can begin with our first identified interval where we go from negative infinity to the negative square root of 57 plus five all divided by two. And now for a test value, we can choose, let's say negative two since it fits within our interval here. And now we need to find F of negative two. Well, F of negative two gives us the negative, the quantity of negative two squared minus five times negative two minus eight. And simplifying we get positive six. Well, since we obtained a positive value from F of negative two, we see that F of X is going to be greater than zero for all X values within this interval where again, we went from negative infinity to the negative square root of 57 plus five all divided by two. So now we can move on to testing our second interval where we went from the negative square root of 57 plus five all divided by two to positive square root of 57 plus five all divided by two. So now again, we need to choose a test value that fits within this interval. So we can just choose zero. And now we need to find F of zero. Well, F of zero gives us zero squared minus five times zero minus eight and simply fine, we just get negative eight. And since we obtained a negative value from F of X, we know that F of X is going to be less than zero for all X values that are within this interval from the negative square root of 57 plus five all divided by two to the positive square root of 57 plus five all divided by two. And now, from here, we just need to test our last interval where we go from the positive square root of 57 plus five all divided by two to positive infinity. And now again, we need to choose a test value this time. Let's choose seven. And it's still fits within our interval here. So we can move forward with substituting seven into F of X. So now we find F of seven. So we have seven squared minus five times seven minus eight and simplifying we get positive six. And since we obtained a positive value, we can conclude that F of X is going to be greater than zero for all X values within this interval from the positive square root of 57 plus five all divided by two to positive infinity. And so now looking at our three conclusions here that we found, we can see that F of X is going to be greater than zero for all X values for all X values from the, from negative infinity to the negative square root of 57 plus five all divided by two. And from, From the positive square root of 57-plus 5 all divided by 2 to positive infinity. Now, from our original inequality, we know that we were looking for when F of X is greater than, or equal to zero for all of the X values. So that means we need to look at the two intervals that we found are greater than zero and adjust what they are closed by. So that they also include being equal to zero. So we can begin with our first interval where we go from negative infinity to the negative square root of 57 plus five all divided by two. And we know that infinity is always excluded. So it will get closed by parentheses. But the other value of the negative square root of 57 plus five all divided by two. Well, this value causes our expression to be equal to zero. So we now need to close it by a bracket. And now we can move on to our second interval where we went from the positive square root of 57 plus five all divided by two to positive infinity. Now again, here, infinity is closed by a parenthesis because it is always excluded. But the first value, the positive square root of 57 plus five five all divided by two now gets closed by a bracket because it causes our expression to be equal to zero. So it is included. So with this, we have found the solution set for our given inequality where our expression is greater than or equal to zero. So with this here we are left with answer choice. B thanks for watching. I hope you found this video helpful and I'll see you again next time.

Solve each inequality. Give the solution set using interval notation. x²-3x ≥ 5

Key Concepts

Inequalities

Quadratic Functions

Interval Notation

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