Solve Quadratic Equationsby Factoring - Example 1
Solve Quadratic Equationsby Graphing - Example 1
Solving Quadratic Equationsby Completingthe Square - Example 1
The Discriminant - Example 1
Adding Subtracting And Multiplying Complex Numbers - Example 2
Liberty University
Solve Quadratic Equations Algebraically - Example 1


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the algebraic Lee we have X Square plus 11 x plus 30 is less than or equal to zero. So let's start off. Unless factor This is though it were in equation, so we would have X plus five and X plus six because five times six is 30 and five plus six is 11 and we're gonna make them equal zero. So that means we're gonna have X plus five equals zero and X plus six equals zero. We're going to subtract five. So X is going to equal negative five, and we're going to subtract six. So X is going to equal negative six. So now that we have that, we can create a number line to determine what are statements actually going to be. So let's create a number line and we've got Right now we have X equals negative five and X equals negative six. So there's your number line now on the number line. We've got three sections we have where X is going to be less than negative six. We have where X is going to be greater than negative five, and then we have the middle section where negative six is less than X, which is less than negative five. So what we want to do is we want to start right here with a number less than negative six. So let's start with X is negative seven. So we're gonna plug that into our inequality. So that would be negative. Seven squared, plus 11 times negative. Seven plus 30. And when I pulled that in, that's going to give me, too. And we're saying two is less than or equal to zero, and that is false. So that is not a solution. So that tells me I need to go to the middle section before I tracked the section over here. So we want a number between negative five and negative six. So let's try. X equals negative 5.5. That's a good number in between. Let's do right in the middle. So we're gonna have negative 5.5 squared, plus 11 times negative 5.5 plus 30. When I plug all that in, I'm going to get negative. 0.25 is less than or equal to zero. That statement is true, So that means that my ex has to be between the negative six and the negative five. So that means my solution is negative. Six is less than or equal to X, which is less center equal to negative five.

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